Topics

According to Michael Faraday we can visualize electric field with the help of lines. These imaginary lines or curves are termed as electric field lines or electric lines of forces. They are drawn such that tangent at any point on an electric line of force gives the direction of the field at that point.

The first figure represents direction of electric field (force on unit positive test charge) due to a positive charge at some points whereas second figure represents corresponding electric field lines.
The number density of lines (number lines crssing unit area normally) represent the relative strength of the field.In other language at points where the intensity is low, the lines of forces will be widely separated and where the intensity is higher, the lines of force will be closely packed.
Another person may draw more lines. But the number of lines is not important. In fact, an infinite number of lines can be drawn in any region. It is the relative density of lines in different regions which is important.

Electric field lines penetrating two surfaces. The magnitude of the field is greater on surface \(A\) than on surface B.

Properties of electric lines of force

(i) Electric lines of forces originate from (+)ve charge and terminate (end) on negative charge i.e. (+)ve charges are sources of electric lines of forces and (-) ve charges are sinks for them.

(ii) Electric lines of forces are open curves (not closed curves like magnetic lines of forces). This property of electric lines of forces signifies the fact that electric field is a conservative force field.
(iii) Two electric lines of force never cross each other because if they do cross then there will be two directions of field at same point.
(iv) The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.
If magnitude as well as direction of electric field is same then electric field is said to be uniform otherwise nonuniform

In the first figure the magnitude as well as direction is uniform. In the second figure magnitude of the electric field is uniform where as direction is non uniform. In the third figure direction of the electric field is uniform where as magnitude is non uniform. In the fourth figure the magnitude as well as direction is non uniform.

The first figure represent uniform electric field where as the remaining three figures represent non uniform electric field

Practice Exercise

Q. 1 The electric field of two point charges separated by 4.14 cm is as shown in figure.
(a) What is the nature of charges?
(b) What is the ratio of magnitude of charges?
(c) Is the field uniform?

(d) Apart from infinity where is the neutral point?
(e) Will a positive charge follow the line of force if free to move?
(f) Where do the extra lines end?

Answers

Q. 1 (a) A is -ve and B is +ve (b) \(\left| q_{A} \right| = 2\left| q_{B} \right|\) (c) Feild is not uniform (d) BC=10 cm (e) No (f) at infinity

When a particle of charge \(q\) and mass \(m\) is placed in an electric field \(E\), the electric force exerted on the charge is \(q\overrightarrow{E}\) according to Equation \(\overrightarrow{E} = \frac{\overrightarrow{F}}{q}\). If this is the only force exerted on the particle, it must be the net force and causes the particle to accelerate according to Newton's second law. Thus,

\[\overrightarrow{a} = \frac{q\overrightarrow{E}}{\text{ }m}\]

If the particle has a positive charge, its force is in the direction of the electric field. If the particle has a negative charge, its acceleration is in the direction opposite the electric field.

Practice Exercise

Q. 1 An electron and a proton are situated in a uniform electric field. What is the ratio of their acceleration?
Q. 2 An infinite plane of positive charge has a surface charge density \(\sigma\). A metal ball \(B\) of mass \(m\) and charge q is attached to a thread and tied to a point A on the sheet PQ . Find the angle \(\theta\) which AB makes with the plane PQ.
Q. 3 A simple pendulum of length \(l\) and bob mass m is hanging in front of a large nonconducting sheet having surface charge density \(\sigma\). If suddenly a charge \(+ q\) is given to the bob & it is released from the position shown in figure. Find the maximum angle through which the string is deflected from vertical.

Q. 4 Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Q. 5 A simple pendulum consists of a small sphere of mass \(m\) suspended by a thread of length 1 . The sphere carries a positive charge q . The pendulum is placed in a uniform electric field of strength E directed vertically upwards. With what period will the pendulum oscillate if the electrostatic force acting on the sphere is less that the gravitational force?Assume the oscillations to be small.
Q. 6 An inclined plane makes an angle of \(30^{\circ}\) with a horizontal is placed in a uniform horizontal electric field \(E = 100{Vm}^{- 1}\). A particle of mass 1 kg and charge 0.01 C , is allowed to slide down from a height of 1 m . If the coefficient of friction is 0.2 , find the time taken by the body to reach the bottom of the inclined plane.
Q. 7 A pendulum bob of mass 80 milligrams and carrying a charge of \(2 \times 10^{- 8}\) coulomb is at equilibrium. in a horizontal uniform electric field of \(20,000{Vm}^{- 1}\). Find the tension in the thread of the pendulum at equilibrium.
Q. 8 Auniform electric field of strength \(10^{6}\text{ }V/m\) is directed vertically downwards. A particle of mass 0.01 kg and charge \(10^{- 6}\) coulomb is suspended by an inextensible thread of length 1 m . The particle is displaced slightly from its mean position and released. Calculate the time period of its oscillation. What minimum velocity should be given to the particle at rest so that it completes full circle in a vertical plane without the thread getting slack? Calculate the maximum and minimum tensions in the thread in this situation.

Area as a Vector

In some calculations it is useful to use area as a vector. Plane surface area or small elemental area of a curved surface can be treated as vector quantities. Its direction is taken along the outward normal drawn on the surface and magnitude is equal to its area. It should be noted here that the direction of outward normal at a given point of a closed surface is unique but in case of open surfaces the terms outward and inward are not well defined and so the choice of outward normal is purely arbitrary.

Flux of a vector field

When a planer area represented by an area vector \(\overrightarrow{A}\) is placed in any uniform vector field (let \(\overrightarrow{X}\) ) then a flux (represented by the symbol \(\Phi\) ) of that vector field is said to be linked through that area and defined as
\[\Phi = \overrightarrow{X} \cdot \overrightarrow{A} \]To understand its meaning clearly let us take example of velocity field. We imagine a hypothetical cylindrical having cross-sectional area A, Lowered in a flowing river in which velocity of water is uniform.

flux of the vector velocity vector on left cross section

\[\Phi_{1} = \overrightarrow{v} \cdot \overrightarrow{\text{ }A} = vAcos180^{\circ} = - vA\text{ }m{\text{ }m}^{3}/sec.\]

flux of the vector velocity vector on right cross section

\[\Phi_{2} = \overrightarrow{v} \cdot \overrightarrow{\text{ }A} = vAcos0^{\circ} = + vA\text{ }m{\text{ }m}^{3}/sec\]

flux of the vector velocity on any elemental area of curved surface

\[\Phi_{3} = \overrightarrow{v} \cdot \overrightarrow{\text{ }A} = vAcos90^{\circ} = 0{\text{ }m}^{3}/sec.\]

As the unit suggests the flux of \(\overrightarrow{V}\) through \(\overrightarrow{A}\) at left end gives the net inflow of water across the left crosssection of the pipe per-sec. Similarly the flux of \(\overrightarrow{v}\) through \(\overrightarrow{A}\) at right end gives the net outflow of water across the right cross-section of the pipe per-sec . Zero flux on curved surface represents there is no flow through curved surface.
Overall we can say that flux of any vector field through an area vector placed in that field defines the net inflow or net outflow of something across that area. (-)ve flux indicates inflow, (+)ve flux indicates outflow.

Flux of electric field

Flux of electric field through an area placed in that field represents the part of the electric field intercepted by that area or the net inflow or outflow of electric lines of forces normally through that area.

When a small area represented by an area vector \(d\overrightarrow{A}\) is placed in any vector field then a flux of that field is said to be linked through that area.

Considering a small surface represented by area vector \(d\overrightarrow{A}\), placed in an electric field \(\overrightarrow{E}\) in such a way that angle between \(\overrightarrow{E}\) and \(d\overrightarrow{A}\) is \(\theta\) then the flux of electric field \(\overrightarrow{E}\) through the area \(d\overrightarrow{A}\) is defined as the dot product of the field vector and that of the area vector and is given by

\[\begin{matrix} & d\Phi = \overrightarrow{E} \cdot \text{ }d\overrightarrow{\text{ }A} = EdAcos\theta \\ \Rightarrow \ & \text{ }d\Phi = (Ecos\theta)dA = \text{~}\text{Product of the area and}\text{~} \end{matrix}\]

that of the component of \(\overrightarrow{E}\) along \(d\overrightarrow{A}\) i.e., along the normal

to the area or,
\(d\Phi = E(dAcos\theta) =\) Product of \(E\) and the projection of the area vector normal to the direction of \(\overrightarrow{E}\).

To calculate the flux of nonuniform electric field through a surface of any arbitrary shape we divide the whole surface into little patches which are so small that over any one patch the surface is practically flat and the electric field doesn't change appreciably from one part of a patch to another. The area of a patch has a certain magnitude and the outward pointing normal to its surface gives its direction. We calculate the fluxes of electric field through each such patches and the algebraic sum of the fluxes through individual patches gives the total flux of electric field through the given surface of arbitrary shape because flux is a scalar quantity.
The flux of nonuniform electric field over an arbitrary surface A is given by the integral

\[\Phi = \int_{}^{}\ d\Phi = \int_{S}^{}\mspace{2mu}\overrightarrow{E} \cdot d\overrightarrow{A} = \int_{S}^{}\mspace{2mu} EdA\widehat{n}\]

where \(\widehat{n}\) is the unit vector in the direction of normal to the surface
In case of closed surface flux is written by special symbol

\[\Phi = \int_{}^{}\ d\Phi = \text{∮}\ \ \overrightarrow{E} \cdot \text{ }d\overrightarrow{\text{ }A} = \text{∮}\ \ EdA\widehat{n}\]

A plane angle is a two dimensional concept. Solid angle is a three dimensional generalization of the two dimensional concept of plane angle. A plane angle is formed at a point by a line segment or an arc but a surface is responsible for the formation of solid angle at a given point. The S.I. unit of plane angle is radian whereas that of solid angle is steradian (Sr).

\(AB\) is the line segment forming plane angle \(\theta\) at O

Plane angle formed by a small arc of a circle at its centre is defined as \(d\theta = \frac{d\mathcal{l}}{r}\)

S is the surface segment forming solid angle \(\Omega\) at O

defined as \(d\theta = \frac{d\mathcal{l}}{r}\)

solid angle formed by a small area of a sphere at its centre is

If a small length element \(d\mathcal{l}\) is oblique such that normal to the length element ( \(\widehat{n}\) ) is making an angle \(\theta\) with radial direction
( \(\widehat{r}\) ) then the plane angle is defined as \(d\theta = \frac{d\mathcal{l}cos\theta}{r}\)

If a small surface element dA is oblique such that normal to the surface element ( \(\widehat{n}\) ) is making an angle \(\theta\) with radial direction
( \(\widehat{r}\) ) then the plane angle is defined as \(d\Omega = \frac{dAcos\theta}{r^{2}}\)

Practice Exercise

Q. 1 A point charge Q is located on the axis of a disc of radius R at a distance a from the plane of the disc. If the soild angle subtended by disc at the point charge is \(\pi/2\), then find the relation between a & R .

Answers

Q. \(1\ a = \frac{R}{\sqrt{3}}\)

Some important results regarding solid angle

(i) A closed surface subtends a solid angle of \(4\pi(Sr)\) at an internal point.

Considering a three dimensional closed surface S of any arbitrary shape. Let O be an interior point. Again, consider a spherical surface of radius ' r ' having its centre at ' O '. Let a small part of the surface, having surface area ' dA ' is contained inside a solid angle \(d\Omega\). This cone emerging from point O encloses an area \({dA}^{'}\) of the spherical surface.
So Solid angle subtended by dA at point \(O = d\Omega =\) solid angle subtended by \(dA^{'}\) at point \(O = \frac{dA^{'}}{r^{2}}\)

Every small area, on the surface \(S\) of any arbitrary shape, may be shown in correspondence with an area on the spherical surface. So, total solid angle subtended by a closed surface of any arbitrary shape at an internal point.

\[\begin{matrix} & \Omega = \int_{}^{}\ d\Omega = \frac{1}{r^{2}}\int_{}^{}\ dA^{'} = \frac{1}{r^{2}}4\pi\left( r^{2} \right) \\ & \ = 4\pi(Sr) \end{matrix}\]

(ii) A closed surface subtends a solid angle of \(0(Sr)\) at an external point.

If the boundary of a hemispherical surface is gradually closed then the solid angle \(\Omega\) start decreasing, when it takes the shape of a closed surface, the solid angle subtended at the external point becomes zero.

Practice Exercise

Q. 1 What is the solid angle subtended by hemisphere (shown in figure) at point P.

Q. 2 Find the solid angle subtended by a cube at its corner

Answers

Q. 1 Zero.
Q. \(2\ \frac{\pi}{2}\)

Practice Exercise

Q. 1 A point charge \(q\) is placed on the axis of a disc (imaginary) of radius \(R\) at a distance \(x\) from the centre of the disc. Find the flux of electric field of the charge \(q\) on disc.
Ans. \(flux\phi = \frac{q}{4\pi\epsilon_{0}}\Omega = \frac{q}{2\epsilon_{0}}(1 - cos\theta) = \frac{q}{2\epsilon_{0}}\left( 1 - \frac{x}{\sqrt{R^{2} + y^{2}}} \right)\)

Answers

Q. \(1\ flux\phi = \frac{q}{4\pi\epsilon_{0}}\Omega = \frac{q}{2\epsilon_{0}}(1 - cos\theta) = \frac{q}{2\epsilon_{0}}\left( 1 - \frac{x}{\sqrt{R^{2} + y^{2}}} \right)\)

The flux of net electric field (due to the charges enclosed by the surface as well as the charges outside it) through any imaginary closed surface(called gaussian surface) of any arbitrary shape is equal to the total charge enclosed by the surface divided by \(\in_{0}\).

\[\text{∮}\ \ {\overrightarrow{E}}_{\text{res}\text{~}} \cdot d\overrightarrow{\text{ }A} = \frac{\Sigma q_{\text{enclosed}\text{~}}}{\epsilon_{0}}\]

Some points to be emphasized about the gauss law :

(i) It is true for any closed surface no matter what its shape and size.
(ii) The \(q\) includes algebraic sum of all charges enclosed by the surface.
(iii) In situations when the surface is so chosen that there are some charges inside and some outside, the \(\overrightarrow{E}\) (whose flux appear in the equation) is due to all charges, just term 'q' in the law represents only total charge inside.
(iv) The gaussian surface should not pass through any discrete charge however, it can pass through a continuous charge distribution.
Gaussian surface should not contain any finite non zero charge.

Whenever a charge is moved in an electrostatic field, work is done by electrostatic forces. An electrostatic field is a conservative force field. Therefore, any work done against the field is stored as potential energy. Change in electric potential energy will be \(\Delta U = - W_{el}\), where \(W_{el}\) is the work done by electrostatic forces.

Electric potential energy between two point charges

Consider two charges \(Q\) and \(q\) separated by a distance \(r_{1}\). If the charge \(q\) is moved along the line joining the charge and the final separation becomes \(r_{2}\), then the work done by the electric force during the process is

\[W_{el} = \int_{r_{1}}^{r_{2}}\mspace{2mu}\frac{kQq}{r^{2}}dr = kQq\left\lbrack \frac{1}{r_{1}} - \frac{1}{r_{2}} \right\rbrack\]

\(\therefore\) The change in potential energy is defined as

\[U_{2} - U_{1} = - W_{el} = kQq\left\lbrack \frac{1}{r_{2}} - \frac{1}{r_{1}} \right\rbrack\]

The potential energy of a two-charge system is taken to be zero, when the distance between the charges is infinity. i.e. \(U = 0\) if \(r = \infty\)
Now, the potential energy of a two charge system when their separation is \(r\), is

\[\begin{matrix} & U(r) - 0 = kQq\left\lbrack \frac{1}{r} - \frac{1}{\infty} \right\rbrack \\ & U(r) = \frac{kQq}{r} = \frac{1}{4\pi\varepsilon_{0}}\frac{Qq}{r} \end{matrix}\]

For like charges U is +ve & for unlike charges U is-ve.

Electric potential energy between more than two point charges

The above equation gives the potential energy of a pair of charges. In case of three charges (say \(q_{1},q_{2}\) and \(q_{3}\) ) there are three pairs \(\left( q_{1},q_{2} \right),\left( q_{2},q_{3} \right)\) and \(\left( q_{3},q_{1} \right)\). Thus the total potential energy of the system will have three terms.

\[U = \frac{kq_{1}q_{2}}{r_{12}} + \frac{kq_{2}q_{3}}{r_{23}} + \frac{kq_{3}q_{1}}{r_{31}}\]

Practice Exercise

Q. 1 Three point charges of 0.1 C each are placed at the corners of an equilateral triangle with side \(L = 1\text{ }m\). If this system is supplied energy at the rate of 1 kW , how much time will be required to move one of the charges on to the mid point of the line joining the two others?
Q. 2 Two fixed, equal, positive charges, each of magnitude \(5 \times 10^{- 5}C\), are located at points \(A\) and \(B\) separated by a distance of 6 m . An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB . The moving charge, when it reaches the point C at a distance of 4 m from O , has kinetic energy of 4 J . Calculate the distance of the farthest point D which the negative charge will reach before returning towards C .

Q. 3 Consider the configuration of a system of four charges each of value +q . Find the work done by external agent in changing the configuration of the system from figure (i) to fig (ii).

Answers

Q. \(1\ 50hrs\).
Q. \(2\ 8.485\text{ }m\)
Q. \(3 - \frac{{kq}^{2}}{a}(3 - \sqrt{2})\)

Electric potential (represented by symbol V) due to a point charge or a charge configuration at a point is defined as

\[V = \frac{U}{q} = \text{~}\text{P.E. per unit test charge.}\text{~}\]

In other words it is amount of work done by external agency to bring a unit positive charge from reference point (usually taken at infinity) to given point

Electric potential due a point charge

In the electric field of a point charge \(Q\) electric potential at a pont will be

\[V(\overrightarrow{r}) = \frac{\frac{kQq}{r}}{q} = \frac{kQ}{r}\]

Potential is a scalar quantity.Electric potential due to a positive charge is taken to be positive and that due to a negative charge is taken to be negative. The potential at a point due to more than one charge can be found simply by adding the potentials due each charge separately.

plot of electric potential

electric potential due to positive charge
electric potential due to negative charge

Practice Exercise

Q. 1 The electric potential at point A is 20 V and at B is -40 V . Find the work done by an external froce and electrostatic force in moving an electron slowly from B to A .
Q. 2 Find the work done by some external force in moving a charge \(q = 2\mu C\) from infinity to a point where electric potential is \(10^{4}\text{ }V\).
Q. 3 Find out the points on the line joining two charges +q and -3 q (kept at a distance of 1.0 m ) where electric potential is zero.
Q. 4 An infinite number of charges each equal to q are placed along the x -axis at \(x = 1,x = 2\), \(x = 8,\ldots\) and so on. Find the potential and the electric field at the point \(x = 0\) due to this set of charges. What will be the potential and electric field if, in the above set up, the consecutive charges have opposite sign?
Q. 5 Four point charges \(+ q, - q, + q\), and -q are placed respectively at the corners \(A,B,C\) and D of a square of side a . (a) Calculate the electric potential at O , the centre of the square. (b) if E and F are the midpoints of sides BC and CD respectively, what is the work that will be done in carrying an electron from : (i) O to E , and (ii) O to F ? Given \(q = 1.0 \times 10^{- 6}C,a = 0.10\text{ }m\) and charge on an electron \(= - 1.6 \times 10^{- 19}C\).

Answers

Q. \(1 - 9.6 \times 10^{- 18}\text{ }J,9.6 \times 10^{- 18}\text{ }J\ \) Q. \(2\ 2 \times 10^{- 2}\text{ }J\)
Q. 3 The potential will be zero at point P on the axis which is either 0.5 m to the left or 0.25 m to the right of charge +q .
Q. \(4\ \frac{2q}{4\pi\epsilon_{0}},\frac{1}{4\pi\epsilon_{0}}\left\lbrack \frac{4}{3}q \right\rbrack;\left\lbrack \frac{2}{3}q \right\rbrack,\frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{4}{5}q \right\rbrack\)
\[\begin{matrix} \text{~}\text{Q.}\text{~}5 & \text{~}\text{(a)}\text{~}0 & \text{~}\text{(b) (i)}\text{~}0 & \text{~}\text{(ii)}\text{~}0 \end{matrix}\]

A charge Q is uniformly distributed over the circumference of a ring. Let us calculate the electric potential at an axial point at a distance \(r\) from the centre of the ring.
The electric potential at P due to the charge element dq of the ring is given by

\[dV = \frac{1}{4\pi\varepsilon_{0}}\frac{dq}{Z} = \frac{1}{4\pi\varepsilon_{0}}\frac{dq}{\left( R^{2} + r^{2} \right)^{1/2}}\]

Hence, the electric potential at P due to the uniformly charged ring is given by

\[\begin{matrix} & V = \int_{}^{}\ \frac{1}{4\pi\varepsilon_{0}}\frac{dq}{\left( R^{2} + r^{2} \right)^{1/2}} = \frac{1}{4\pi\varepsilon_{0}}\frac{1}{\left( R^{2} + r^{2} \right)^{1/2}}\int_{}^{}\ dq \\ & \ = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{\sqrt{\left( R^{2} + r^{2} \right)}} \end{matrix}\]

plot of electric potential

A non-conducting disc of radius ' R ' has a uniform surface charge density \(\sigma C/m^{2}\). Let us calcuate the potential at a point on the axis of the disc at a distance ' \(r\) ' from its centre. The symmemtry of the disc tells us that the appropriate choice of element is a ring of radius x and thickness dx . All points on this ring are at the same distance \(Z = \sqrt{x^{2} + r^{2}}\), from the point \(P\). The charge on the ring is \(dq = \sigma dA = \sigma(2\pi xdx)\) and so the potential due to the ring is

\[dV = \frac{1}{4\pi\varepsilon_{0}}\frac{dq}{Z} = \frac{1}{4\pi\varepsilon_{0}}\frac{\sigma(2\pi xdx)}{\sqrt{x^{2} + r^{2}}}\]

Since potential is scalar
The potential due to the whole disc is given by

\[\begin{matrix} & & V = \frac{\sigma}{2\varepsilon_{0}}\int_{0}^{R}\mspace{2mu}\mspace{2mu}\frac{xdx}{\sqrt{x^{2} + r^{2}}} \\ & & \ = \frac{\sigma}{2\varepsilon_{0}}\left\lbrack \left( x^{2} + r^{2} \right)^{1/2} \right\rbrack_{0}^{R} \\ & \ = \frac{\sigma}{2\varepsilon_{0}}\left\lbrack \left( R^{2} + r^{2} \right)^{1/2} - r \right\rbrack & \text{(xB)} \end{matrix}\]

Let us see this expression at large distance when \(r \gg R\).

\[V = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}\text{, where}\text{~}Q = \pi r^{2}\sigma\text{~}\text{is the total charge on the disc.}\text{~}\]

Thus, we conclude that at large distance, the potential due to the disc is the same as that of a point charge Q.
plot of electric potential

Place a charge \(q\) in electric field \(\overrightarrow{E}\). This field exerts a force \(\overrightarrow{F} = q\overrightarrow{E}\) on the charge. If you displace the charge by by \(d\overrightarrow{r}\) then field will do some work \(dW = \overrightarrow{F} \cdot d\overrightarrow{r} = q\overrightarrow{E} \cdot d\overrightarrow{r}\) on the charge.
Change in electric potential energy during this displacement will be given by

\[dU = - q\overrightarrow{E} \cdot \text{ }d\overrightarrow{r}\]

Therfore potential difference between initial and final point will be given by

\[dV = - \overrightarrow{E} \cdot \overrightarrow{dr}\]

Calculation of electric potential difference from electric field

\[\begin{matrix} & dV = - \overrightarrow{E} \cdot \overrightarrow{dr} \\ & \ \Rightarrow \int_{V\left( {\overrightarrow{r}}_{1} \right)}^{V\left( {\overrightarrow{r}}_{2} \right)}\mspace{2mu}\mspace{2mu} dV = - \int_{{\overrightarrow{r}}_{1}}^{{\overrightarrow{r}}_{2}}\mspace{2mu}\mspace{2mu}\overrightarrow{E} \cdot \text{ }d\overrightarrow{r} \\ & \ \Rightarrow \text{ }V\left( \overrightarrow{r_{2}} \right) - V\left( \overrightarrow{r_{1}} \right) = \int_{}^{}\ dV = - \int_{{\overrightarrow{r}}_{1}}^{{\overrightarrow{r}}_{2}}\mspace{2mu}\mspace{2mu}\overrightarrow{E} \cdot \text{ }d\overrightarrow{r} \end{matrix}\]

In cartesian coordinate system

\[\begin{matrix} d\overrightarrow{r} = dx\widehat{i} + dy\widehat{j} + dz\widehat{k} \\ \Rightarrow \overrightarrow{E} \cdot d\overrightarrow{r} = E_{x}dx + E_{y}dy + E_{z}dz \\ \Rightarrow V\left( \overrightarrow{r_{2}} \right) - V\left( \overrightarrow{r_{1}} \right) = - \int_{r_{1}}^{r_{2}}\mspace{2mu}\mspace{2mu}\overrightarrow{E} \cdot d\overrightarrow{r} = - \int_{x_{1}}^{x_{2}}\mspace{2mu}\mspace{2mu} E_{x}dx - \int_{y_{1}}^{y_{2}}\mspace{2mu}\mspace{2mu} E_{y}dy - \int_{z_{1}}^{z_{2}}\mspace{2mu}\mspace{2mu} E_{z}dz \end{matrix}\]

Illuatratrion:

Electric field in a region is given by \(\overrightarrow{\mathbf{E}} = (\mathbf{2}\widehat{\mathbf{i}} + \mathbf{3}\widehat{\mathbf{j}} - \mathbf{4}\widehat{\mathbf{k}})V/m\). Find the potential difference between points ( \(0,0,0\) ) and (1,2,3) in this region

Sol. p.d. across the points
\[V = \int_{{\overrightarrow{r}}_{1}}^{{\overrightarrow{r}}_{2}}\mspace{2mu}\overrightarrow{E}.d\overrightarrow{r} = - \int_{x_{1}}^{x_{2}}\mspace{2mu}{\overrightarrow{E}}_{x}dx - \int_{y_{1}}^{y_{2}}\mspace{2mu} E_{y}dy - \int_{z_{1}}^{z_{2}}\mspace{2mu} E_{z}dz = - \int_{0}^{1}\mspace{2mu} 2dx + \int_{0}^{2}\mspace{2mu} 3dy + \int_{0}^{3}\mspace{2mu} - 4dx \]\(= - 2 - 6 + 12 = 4\) volts.

Practice Exercise

Q. 1 An electric field of \(20\text{ }N/C\) exists along the X -axis in space. Calculate the potential difference \(V_{A} - V_{B}\) where the points \(A\) and \(B\) are given by, \(A = (0,0);B = (4\text{ }m,2\text{ }m)\).
Q. 2 In figure, two points A and B are located within a region in which there is an electric field.
(i) The sign of potential difference \(\Delta V = V_{B} - V_{A}\) is
(ii) A negative charge is placed at A and then moved to B . The sign of change in potential energy of the charge-field system for this process is

Q. 3 Two points A and B are 2 cm apart and a uniform electric field E acts along the straight line AB directed from A to B with E=200 N/C. A particle of charge \(+ 10^{- 6}C\) is taken from A to B along AB. Calculate (a) the force on the charge (b) the potential difference \(V_{A} - V_{B}\) and (c) the work done on the charge by \(\overrightarrow{\mathbf{E}}\).

Answers

Q. \(1 - 80\text{ }V\)
Q. 2
(i) -ve
(ii) \(\ +\) ve
Q. \(32 \times 10^{- 4}\text{ }N,4\) volt and \(4 \times 10^{- 6}\text{ }J\)

For our convenience we select potential of a point to be zero. This point is called reference point. Usually reference point is taken at infinity. In the above equation let us take \(\overrightarrow{r_{1}} = \infty\) and \(\overrightarrow{r_{2}} = \overrightarrow{r}\), then the potential at a point can be written as

\[V(\overrightarrow{r}) = - \int_{\infty}^{\overrightarrow{r}}\mspace{2mu}\overrightarrow{E} \cdot d\overrightarrow{r}\]

Practice Exercise

Q. 1 An electric field \(\overrightarrow{E} = \overrightarrow{i}\) Ax exists in the space, where \(A = 10\text{ }V/m^{2}\). Take the potential at ( \(10\text{ }m,20\text{ }m\) ) to be zero. Find the potential at the origin.

Answers

Q. 1500 V

\[\begin{matrix} dV = & \ - \overrightarrow{E} \cdot d\overrightarrow{r} \\ dV = & \ - (Ecos\theta)dr \\ & \ - \frac{dV}{dr} = Ecos\theta \end{matrix}\]

(i.e., Rate of decrease of potential)

Where \(Ecos\theta\) is component of field in the direction of displacement. From the above expression.
Potential decreases maximum in the direction of field, \(\theta = 0^{\circ}\). It is also clear that, \(- \frac{dV}{dr}\) is maximum in the direction of the field, so, we may conclude that, the electric potential decreases at maximum rate in the direction of the field.
The cartesian component of electric field can be written as

\[\overrightarrow{E} = E_{x}\widehat{i} + E_{y}\widehat{j} + E_{z}\widehat{k};\]

and an infinitesimal displacement is \(d\overrightarrow{r} = dx\widehat{i} + dy\widehat{j} + dz\widehat{k}\)
Thus,

\[\begin{matrix} & dV = - \overrightarrow{E} \cdot d\overrightarrow{r} \\ & \ = - \left\lbrack E_{x}dx + E_{y}dy + E_{z}dz \right\rbrack \end{matrix}\]

for a displacement in the x -direction,

\[\begin{matrix} & dy = dz = 0\text{~}\text{and so}\text{~} \\ & dV = - E_{x}dx.\text{~}\text{Therefore,}\text{~} \\ & E_{x} = - \left( \frac{dV}{dx} \right)_{y,\text{~}\text{zonstant}\text{~}} \end{matrix}\]

A derivative in which all variables except one are held constant is called partial derivative and is written with \(\partial\) instead of d . The electric field is, therefore,

\[\left. \ \begin{matrix} E_{x} = - \frac{\partial V}{\partial x} \\ E_{y} = - \frac{\partial V}{\partial y} \\ E_{z} = - \frac{\partial V}{\partial z} \end{matrix} \right\}\]

i.e., if scalar potential function is given field can be calculated taking help of the above relations as

\[\overrightarrow{E} = - \left\lbrack \frac{\partial V}{\partial x}\widehat{i} + \frac{\partial V}{\partial y}\widehat{j} + \frac{\partial V}{\partial z}\widehat{k} \right\rbrack\ |\overrightarrow{E}| = \sqrt{\left( \frac{\partial v}{\partial x} \right)^{2} + \left( \frac{\partial v}{\partial y} \right)^{2} + \left( \frac{\partial v}{\partial z} \right)^{2}}\]

Practice Exercise

Q. 1 The electric potential existing in space is \(V(x,y,z) = A(xy, + yz + zx)\). (a) Write the dimensional formula of A. (b) Find the expression for the electric field. (c) If A is 10 IS units, find the magnitude of the electric field at ( \(1\text{ }m,1\text{ }m,1\text{ }m\) ).

Answers

Q. 1
(a) \({MT}^{- 3}I^{- 1}\)
(b) \(- A\{\overrightarrow{i}(y + z) + \overrightarrow{j}(z + x) + \overrightarrow{k}(x + y)\}\)
(c) \(35\text{ }N/C\)

A locus of points in space that all have the same potential is called an equipotential surface.

Properties of equipotential surfaces :

(i) Work done in moving a charge over an equaipotential surface is zero. Thus the work done in moving a charge +q from one point A to another point B on a equipotential surface is given by;

\[W_{AB} = - q\left( {\text{ }V}_{B} - V_{A} \right) = - q(0) = 0\]

(ii) The electric field is always perpendicular to an equipotential surface. Referring to figure, we have,

\[dV = - \overrightarrow{E} \cdot \text{ }d\overrightarrow{l}\]

Since \(dV = 0\) for an equipotential surface,

\[\therefore\ \overrightarrow{E} \cdot \text{ }d\overrightarrow{l} = 0\]

This means that \(\overrightarrow{E}\) is perpendicular to \(d\overrightarrow{l}\).

In other words, electric field (or electric lines of force) are perpendicular to the equipotential surface.

equipotential surface of uniform electric field

equipotential surface of electric field of charges \(+ q\) and \(+ q\)

equipotential surface of electric field of a charge +q

equipotential surface of electric field of charges \(+ q\) and \(+ q\)

(iii) The spacing between equipotential surfaces enables us to identify regions of strong and weak field.

We know that \(E = - \frac{dV}{dr}\)
For a given dV (i.e., constant dV ), \(E \propto 1/dr\). This means that where the equipotential surfaces are crowded, the electric field intensity is greater and vice-versa.
In the above figure equipotential surfaces having constant potrntial difference between two consecutive surfaces are shown. Near the point charge equipotential surfaces are crowded, the electric field intensity is greater
(iv) Two equipotential surfaces can be never intersect. If two equipotential surfaces could intersect, then at the point of intersection there would be two values of electric potential which is not possible.

Example:

(i) In the field of a point charge, the equipotential surfaces are spheres centered on the point charge.
(ii) In a uniform electric field, the equipotential surfaces are planes which are perpendicular to the fields lines.
(iii) In the fields of an infinite line charge, the equipotential surfaces are co-axial cylinders having their axes at the line charge.

If the charge on the shell \(= q\)
(i) for \(r > R\)

\[E = \frac{kq}{r^{2}}\]

\[\Rightarrow \ V = - \int_{\infty}^{r}\mspace{2mu} Edr = - \int_{\infty}^{r}\mspace{2mu}\frac{kq}{r^{2}}dr = \frac{kq}{r} \]

(ii) for \(r = R,V = \frac{kq}{R}\)
(iii) for \(r < R\)
\[E = 0(r < R)\]

\[\Rightarrow \ V = - \int_{\infty}^{r}\mspace{2mu} Edr = - \left\lbrack \int_{\infty}^{R}\mspace{2mu}\mspace{2mu} Edr + \int_{R}^{r}\mspace{2mu}\mspace{2mu} Edr \right\rbrack = - \int_{\infty}^{R}\mspace{2mu}\frac{kq}{r^{2}}dr - \int_{R}^{r}\mspace{2mu} 0dr = \frac{kq}{R}\]

plot of electric potential

Electric Potential due to a uniformly charged spherical volume

If the total charge \(= Q\)
(i) for \(r > R\),

Volume charge density \(\rho = \frac{Q}{\frac{4}{3}\pi R^{3}}\)

\[\begin{matrix} E(r > R) = \frac{kQ}{r^{2}} \\ V = - \int_{\infty}^{}\mspace{2mu}\mspace{2mu}\frac{kQ}{r^{2}}dr = \frac{kQ}{r} \end{matrix}\]

(ii) for \(r = R,V = \frac{kQ}{R}\)
(iii) \(r < R,E(r < R) = \frac{\rho r}{3\varepsilon_{o}} = \frac{Q.r}{\frac{4}{3}\pi R^{3}.3\varepsilon_{o}}\)

\[\begin{matrix} \Rightarrow \ & E(r < R) = \frac{Qr}{4\pi\varepsilon_{o}R^{3}} = \frac{kQr}{R^{3}} \\ & \text{ }V = - \left\lbrack \int_{\infty}^{R}\mspace{2mu}\mspace{2mu} Edr + \int_{R}^{r}\mspace{2mu}\mspace{2mu} Edr \right\rbrack = - \int_{\infty}^{R}\mspace{2mu}\mspace{2mu}\frac{kQ}{r^{2}}dr - \int_{R}^{r}\mspace{2mu}\mspace{2mu}\frac{kQr}{R^{3}}dr = \frac{kQ}{2R}\left\lbrack 3 - \frac{r^{2}}{R^{2}} \right\rbrack \end{matrix}\]

plot of electric potential

Practice Exercise

Q. 1 Find the potential difference between to points for a infinite sheet having uniform surface charge density \(\sigma\)
Q. 2 Find the work done by external agent to bring a charge q from centre of shell to centre of spherical charged volume

Q. 3 A circular ring of radius \(R\) with uniform positive charge density \(\lambda\) per unit length is located in the \(y - z\) plane with its centre at the origin \(O\). A particle of mass \(m\) and positive charge \(q\) is projected from the point \(P\lbrack\sqrt{3}R,0,0\rbrack\) on the positive x -axis directly towards O , with initial speed v . Find the smallest (non zero) value of the speed such that the particle does not return to \(P\).

Answers

Q. \(1\ {\text{ }V}_{12} = \frac{\sigma}{2\varepsilon_{0}}\left( r_{2} - r_{1} \right)\)
Q. \(2\frac{Qq}{8\pi\varepsilon_{0}R}\)
Q. \(3\sqrt{\frac{\lambda q}{2\epsilon_{0}m}}\)

\[U = \frac{1}{2}\sum_{i = 1}^{N}\mspace{2mu}{\text{ }V}_{i}q_{i}\]

Where
\(q_{i} =\) charge on i-th particle and
\(V_{i} =\) potential at the location of i-th particle due to remaining charge particles

Energy Density of an Electric Field :

Work is done in creating any electrostatic system. This work is stored as energy in the field. The energy per unit volume or the energy density \(U\), of the field is given as

\[U = \frac{1}{2}\text{ }K\varepsilon_{0}E^{2}\]

Practice Exercise

Q. 1 Two uniformly charged spherical shells have charges \(q_{1}\) and \(q_{2}\) and radii \(r_{1}\) and \(r_{2}\). Find the potential energy of the system if their cerntres are separated by a distnace \(r\).

Answers

Q. \(1\frac{q_{1}^{2}}{8\pi\varepsilon_{0}r_{1}} + \frac{q_{2}^{2}}{8\pi\varepsilon_{0}r_{2}} + \frac{q_{1}q_{2}}{4\pi\varepsilon_{0}r}\)

If a dipole is placed in a uniform electric field E ,
Force on the dipole is zero.
Torque on the dipole is given as

Here the net force is zero, but there can be a torque.
\[\tau = (qE)(2lsin\theta) = pEsin\theta \]In vector form \(\overrightarrow{\tau} = \overrightarrow{p} \times \overrightarrow{E}\)
This torque has a tendency to orient dipole moment vector in the direction of field

Potential energy due to action of electric field

\[\begin{matrix} & \tau = pEcos\alpha \\ & \ \therefore{\text{ }W}_{fd} = \int_{}^{}\ dW = pE\int_{0}^{90 - \theta}\mspace{2mu}\mspace{2mu} cos\alpha d\alpha = pE\lbrack sin\alpha\rbrack_{0}^{90 - \theta} = pEcos\theta \\ & \ \therefore U_{\theta} = - pEcos\theta = - \overrightarrow{p} \cdot \overrightarrow{E}\ \left( \text{~}\text{taking}\text{~}U = 0\text{~}\text{at}\text{~}\theta = 90^{\circ} \right) \end{matrix}\]

When \(\theta = 0^{0}\), the dipole moment p is in the direction of the field E and the dipole is in stable equilibrium. If it is slightly displaced, it performs oscillations.
When \(\theta = 180^{\circ}\), the dipole moment p is opposite to the direction of the field E and the dipole is in unstable equilibrium.

Practice Exercise

Q. 1 A dipole of dipole moment p is placed at origin along x -axis. Another dipole of dipole moment p is kept at \((0,1,0)\) along \(y\)-axis. Find the resultant potential and electric field at \((1,0,0)\)
Q. 2 Due to electric dipole, electric field at a distance r on axial position is \({\overrightarrow{E}}_{1}\) and at distance r on equatorial position is \({\overrightarrow{E}}_{2}\). What is the relation between \({\overrightarrow{E}}_{2}\) and \({\overrightarrow{E}}_{1}\).
Q. 3 Three charges Q, Q and - 2Q are placed at the three corners of an equilateral triangle of side a. Find the dipole moment of the combination.

Answers

Q. \(1\ v = kp\left( \frac{2\sqrt{2} - 1}{2\sqrt{2}} \right),\overrightarrow{E} = kp\left\lbrack \frac{(8\sqrt{2} - 3)}{4\sqrt{2}}\widehat{i} + \frac{1}{4\sqrt{2}}\widehat{j} \right\rbrack\)
Q. \(2\ {\overrightarrow{E}}_{1} = - 2{\overrightarrow{E}}_{2}\)
Q. \(3p = \sqrt{3}qa\)

Dipole in an external non uniform electric field :

If a dipole is placed in non uniform electric field then let electric field at the location of positive charge will be \({\overrightarrow{E}}_{1}\) and at the location of -ve charge be \({\overrightarrow{E}}_{2}\). Usually for non uniform electric field \({\overrightarrow{E}}_{1} + {\overrightarrow{E}}_{2}\) hence dipole will experience a net force (usually) which is equal to

\[\begin{matrix} & {\overrightarrow{F}}_{\text{net}\text{~}} = q{\overrightarrow{E}}_{1} + ( - q){\overrightarrow{E}}_{2} \\ = & q\left( {\overrightarrow{E}}_{1} - {\overrightarrow{E}}_{2} \right) \\ = & \ - \frac{q\left( {\overrightarrow{E}}_{2} - {\overrightarrow{E}}_{1} \right)}{\mathcal{l}} = - p\frac{\partial\overrightarrow{E}}{\partial\mathcal{l}} \end{matrix}\]

Where \(\frac{\partial\overrightarrow{E}}{\partial\mathcal{l}} =\) rate of change of electric field in the direction of dipole moment. Torque of this electric field about geometric centre of dipole is still given by

\[\overrightarrow{\tau} = \overrightarrow{p} \times \overrightarrow{E}\]

In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal. The free electrons form a kind of 'gas'; they collide with each other and with the ions, and move randomly in different directions. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions.

Inside a conductor, electrostatic field is zero

Consider a conductor, neutral or charged. There may also be an external electrostatic field. In the static situation, when there is no current inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor. This fact can be taken as the defining property of a conductor. A conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift. In the static situation, the free charges have so distributed themselves that the electric field is zero everywhere inside. Electrostatic field is zero inside a conductor.

Further Explanation

What happens if conductor is placed in an extermal field.
Lets keep a positive charge \(q\) near a neutral or a charged conductor then the electrons goes close to \(q\). The redistribution of electrons inside conductor takes place which generates an internal electric field \({\overrightarrow{E}}_{\text{int}\text{~}}\).

neutral conductor

So an \(e^{-}\)experience \({\overrightarrow{E}}_{\text{net}\text{~}}\)
If \(E_{\text{int}\text{~}} \neq E_{\text{ext}\text{~}}\), then the \(e^{-}\)move such that they will create a stronger \({\overrightarrow{E}}_{\text{int}\text{~}}\) which will tend to cancel \(E_{\text{ext}\text{~}}\) This constitutes current and therefore energy conservation is not valid.
\({\overrightarrow{E}}_{\text{net}\text{~}}\) has to be O instantaneously.

\[{\overrightarrow{E}}_{net} = {\overrightarrow{E}}_{ext} + {\overrightarrow{E}}_{int} = 0\]

The interior of a conductor can have no excess charge in the static situation

A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. When the conductor is charged, the excess charge can reside only on the surface in the static situation.

Explanation

This follows from the Gauss's law. Consider any arbitrary volume element \(v\) inside a conductor. If we consider any small gaussian surface inside

\[\text{∮}\ \ \overrightarrow{E} \cdot \text{ }d\overrightarrow{\text{ }A} = 0\ \lbrack\text{~}\text{as}\text{~}E = 0\rbrack\]

On the closed surface \(S\) bounding the volume element \(v\), electrostatic field is zero. Thus the total electric flux through \(S\) is zero. Hence, by Gauss's law, there is no net charge enclosed by \(S\).

\[\Rightarrow \ q_{\text{enclosed}\text{~}} = 0\]

Since the surface \(S\) can be made as small as you like, i.e., the volume \(v\) can be made vanishingly small. This means there is no net charge at any point inside the conductor, and any excess charge must reside at the surface.

Note : Thus Solid " conducting" sphere is same as a shell.
Note : You may emphasise again but \(q_{\text{in}\text{~}} = 0\) does not imply that \(E = 0\) from gauss law.

At the surface of a charged conductor, electrostatic field must be normal to the surface at every point

If \(\mathbf{E}\) were not normal to the surface, it would have some non-zero component along the surface. Free charges on the surface of the conductor would then experience force and move. In the static situation, therefore, \(\mathbf{E}\) should have no tangential component. Thus electrostatic field at the surface of a charged conductor must be normal to the surface at every point. (For a conductor without any surface charge density, field is zero even at the surface.)

Cavity inside a conductor containing no charge inside it

Cavity is a place surrounded from all sides by the conductor such that without touching the body we can't reach cavity.

Electrostatic shielding

Consider a conductor with a cavity, with no charges inside the cavity. A remarkable result is that the electric field inside the cavity is zero, whatever be the size and shape of the cavity and whatever be the charge on the conductor and the external fields in which it might be placed. We have proved a simple case of this result already: the electric field inside a charged spherical shell is zero. The proof of the result for the shell makes use ofthe spherical symmetry of the shell. But the vanishing of electric field in the (charge-free) cavity of a conductor is, as mentioned above, is a general result. A related result is that even if the conductor is charged or charges are induced on a neutral conductor by an external field, all charges reside only on the outer surface of a conductor with cavity.
Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence: thefield inside the cavity is always zero. This is known as electrostatic shielding. The effect can be made use of in protecting sensitive instruments from outside electrical influence. Figure gives a summary of the important electrostatic properties of a conductor.

If there is no charge present inside the cavity then field inside it is zero.

(This does not mean that we are implying "no net charge". If there is some charge present here and there in cavity such that there sum total is zero then the following disscussion will not be valid)
Let us consider a gauesion surface just near to cavity.

\[\begin{matrix} \therefore & {\text{∮}\ }_{E} \cdot \text{ }d\overrightarrow{\text{ }A} = 0 \\ \Rightarrow & q_{\text{in}\text{~}} = 0 \end{matrix}\]

From this we don't have proved that no charge resides on cavity but have proved that net charge on cavity surface is O .
Now suppose a conductor of arbitrary shape contains a cavity as shown in figure.

Let us assume that no charges are inside the cavity. In this case, the electric field inside the cavity must be zero regardless of the charge distribution on the outside surface of the conductor. Furthermore, the field in the cavity is zero even if an electric field exists outside the conductor.
To prove this point, we use the fact that every point on the conductor is at the same electric potential, and therefore any two points \(A\) and \(B\) on the surface of the cavity must be at the same potential. Now imagine that a field \(E\) exists in the cavity and evaluate the potential difference \(V_{B} - V_{A}\) defined by equation.

\[\begin{matrix} & \Delta V \equiv \frac{\Delta U}{q_{0}} = - \int_{A}^{B}\mspace{2mu}\mspace{2mu} E \cdot ds \\ & {\text{ }V}_{B} - V_{A} = - \int_{A}^{B}\mspace{2mu}\mspace{2mu} E \cdot ds \end{matrix}\]

Because \(V_{B} - V_{A} = 0\), the integral of \(E \cdot d\) s must be zero for all paths between any two points \(A\) and \(B\) on the conductor. The only way that this can be true for all paths is if E is zero everywhere in the cavity.

Thus, we conclude that a cavity surrounded by conducting walls is a field-free region as along no charges are inside the cavity.

Cavity inside a conductor containing charge inside it

1. Equal and opposite charge is induced on the inner surface of cavity.

Figure shows a conductor with a cavity inside it. A charge \(q\) is placed inside cavity.
In electrostatic equilibrium charge distribution will be as shown in figure
charge on inner surface of cavity is \(- q\) since material of conductor is initially neutral, equal and opposite charge appears on outer surface.

Let us consider a gaussian surface just outside cavity inside material of conductor. As \(\overrightarrow{E}\) in material of conductor is zero.

\[\text{∮}\ \ E \cdot dA = 0\ \therefore\ q_{\text{enclosed}\text{~}} = 0\]

Thus equal and opposite charge is induced on the inner surface of cavity.
2. The electric field due to charges on the inner surface of conductor nullifies the electric field of point charge all the points outside the inner surface
3. The electric field due to charges on the outer surface of conductor is zero for all the points inside the outer surface seperately
Consider a charged conductor having charge \(+ q_{1}\) and Q is kept inside the cavity. Lets call charge Q inside cavity as \(A\), the induced charge - \(Q\) on the surface of the cavity as \(B\) and the charge on the surface of the conductor \(Q + q_{1}\) at C .

Now field inside the conductor is, \({\overrightarrow{E}}_{\text{net}\text{~}}\) and \({\overrightarrow{E}}_{A},{\overrightarrow{E}}_{B}\) and \({\overrightarrow{E}}_{C}\) are fields due to charge \(A,B\) and C inside the conductor.
and, \({\overrightarrow{E}}_{P,\text{~}\text{net}\text{~}} = {\overrightarrow{E}}_{A} + {\overrightarrow{E}}_{B} + {\overrightarrow{E}}_{C} = 0\)

Now the electric field due to charges on the outer surface of conductor is zero for all the points inside the conductor seperately and the \({\overrightarrow{E}}_{B} + {\overrightarrow{E}}_{A}\) is zero seperately.
Special case: When spherical cavity in present inside spherical conductor and charge is at the centre. Then field inside the cavity is just due to charge \(A\) only as new field lines are already \(\bot\) to \(B\) and \(E\) due to \(B = 0\). [Symmetrical distribution]

Note : When if we displace the charge \(q\) inside the cavity, it will only affect the charge distribution at \(B\) the charge distribution at C will remain unaffected.

Electrostatic Pressure

If a small piece of radius \(b\) is removed from a charged spherical shell of radius \(a( \gg b)\), calculate electric intensity at the midpoint of the aperture, assuming the density of charge to be \(\sigma\).

Consider the shell to be made up of a disc of radius \(b\) and the remainder. If \(E_{D}\) and \(E_{R}\) are the intensities due to disc and the remainder respectively at P , then for a charged spherical shell (or conductor)

\[E_{\text{out}\text{~}} = \frac{\sigma}{\varepsilon_{0}}\text{~}\text{and}\text{~}E_{\text{in}\text{~}} = 0\]

Now as for outside the shell both \(E_{D}\) and \(E_{R}\) will be directed outwards while inside \(E_{R}\) will be outwards while \(E_{D}\) inwards so that :

\[\begin{array}{r} E_{\text{out}\text{~}} = E_{R} + E_{D}\text{~}\text{and}\text{~}E_{\text{in}\text{~}} = E_{R} - E_{D}\#(2) \end{array}\]

And hence equating Eqs. (1) and (2),

\[E_{R} + E_{D} = \frac{\sigma}{\varepsilon_{0}}\text{~}\text{and}\text{~}E_{R} - E_{D} = 0\]

Solving these for \(E_{R}\) and \(E_{D}\) :

\[E_{R} = E_{D} = \frac{\sigma}{2\varepsilon_{0}}\]

i.e., field at the aperture will be ( \(\sigma/2\varepsilon_{0}\) ) directed outwards.
As intensity on the disc (element) the to remainder is \(\left( \sigma/2\varepsilon_{0} \right)\), electric force on it will be,

\[dF = dqE = (\sigma ds)\left\lbrack \frac{\sigma}{2\varepsilon_{0}} \right\rbrack = \left\lbrack \frac{\sigma^{2}}{2\varepsilon_{0}} \right\rbrack ds\]

So force per unit area on a charged conductor due to its own charge

\[\frac{dF}{ds} = \frac{\sigma^{2}}{2\varepsilon_{0}} = \frac{1}{2}\varepsilon_{0}E^{2}\ \left\lbrack \text{~}\text{as for a conductor}\text{~}E = \frac{\sigma}{\varepsilon_{0}} \right\rbrack\]

This force is called or electrostatic pressure. ]

Important :

(i) Thus facing surfaces have equal and opposite charges.
(ii) Using this also prove that outer faces of the two last plates have equal charges.

Practice Exercise

Q. 1 A hollow, uncharged spherical conductor has inner radius a and outer radius b. A positive point charge +q is in the cavity at the centre of the sphere. Make the graph E and potential \(V(r)\) everywhere, assuming that \(V = 0\) at \(r = \infty\).
Q. 2 A solid conducting sphere having a charge \(Q\) is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of hollow shell be V. What will be the new potential difference between the same two surfaces if the shell given a charge -3 Q ?

Q. 3 Two conducting plates X and Y , each having large surface area A (on one side), are placed parallel to each other as shown in the figure. The plate X is given a charge Q whereas the other is neutral. Find (a) the surface charge density at the inner surface of the plate \(X\), (b) the electric field at a point to the left of the plates, (c) the electric field at a point in between the plates and (d) the electric filed at a point to the right of the plates.

Answers

Q. 1

Q. 2 V
Q. \(3\ \) (a) \(\frac{Q}{2\text{ }A}\ \) (b) \(\frac{Q}{2\text{ }A\varepsilon_{0}}\) towards left (c) \(\frac{Q}{2\text{ }A\varepsilon_{0}}\) towards right (d) \(\frac{Q}{2\text{ }A\varepsilon_{0}}\) towards right

Q. 1 A ring has charge \(Q\) and radius \(R\). If a charge \(q\) is placed at its center, then calculate the increase in tension in the ring.
Sol. Let us take an elementry part subtending an angle do at centre.
Charge on the elementry part will be
\[dQ = \frac{Q}{2\pi R}(Rd\theta) = \frac{Q}{2\pi}\text{ }d\theta \]Free body diagram will be

For qeuilibrium \(2\text{ }Tsin\frac{\text{ }d\theta}{2} = \frac{kQq}{2\pi R^{2}}\text{ }d\theta\)
\(\Rightarrow 2\text{ }T\frac{\text{ }d\theta}{2} = \frac{kQq}{2\pi R^{2}}\text{ }d\theta\ \lbrack\because\) for small \(\theta,sin\theta \approx \theta\rbrack\)
\(\therefore T = \frac{kQq}{2\pi R^{2}}\) Ans.
Q. 2 Arigid insulated wire frame in the form of a right-angled triangle ABC , is set in a vertical plane as shown in Fig. Two beads of equal masses \(m\) each and carrying charges \(q_{1}\) and \(q_{2}\) are connected by a cord of length 1 and can slide without friction on the wires

Considering the case when the beads are stationary, determine
(a) the angle \(\alpha\) (b) the tension in the cord and (c) the normal reaction on the beads.

If the cord is now cut what are the value of the charges for which the beads continue to remain stationary?
Sol. Tension and electrostatic force are in opposite direction and along the string. Now each bead is in equilibrium under three concurrent forces.
(i) Normal reaction ( N )
(ii) Weight (mg) and
(iii) \(T - F_{c}\)
where \(F_{e} = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{q_{1}q_{2}}{l^{2}}\)
Applying Lami's theorem for both beads.

\[\begin{array}{r} \frac{N_{1}}{sin(120 - \alpha)} = \frac{mg}{cos\alpha} = \frac{T - F_{e}}{cos60^{\circ}}\#(i) \end{array}\]

\[\begin{array}{r} \frac{N_{2}}{sin\left( 60^{\circ} + \alpha \right)} = \frac{mg}{sin\alpha} = \frac{T - F_{e}}{cos30^{\circ}}\#(ii) \end{array}\]

Dividing equation (i) by (ii), we have

\[tan\alpha = \frac{cos30^{\circ}}{cos60^{\circ}} = \sqrt{3}\ \therefore\alpha = 60^{\circ}\]

\[\begin{array}{r} T = F_{e} + mg = \left( \frac{1}{4\pi\varepsilon_{0}} \right)\frac{q_{1}q_{2}}{l^{2}} + mg\#(iii) \end{array}\]

\(N_{1} = \sqrt{3}mg\) and \(N_{2} = mg\)
From Eq. (iii) \(T = 0\) when string is cut
or \(\ q_{1}q_{2} = - \left( 4\pi\varepsilon_{0} \right)mgl^{2}\)
Q. 3 A positive charge \(Q\) is distributed uniformly over the circumference of a thin circular ring of radius a metres. Calculate the force on a positive point charge q
(i) if it was placed at a distance x from the centre of the ring along its axis.
(ii) if it was placed at the centre of ring.
(iii) At what point on the axis will the force be maximum.
(iv) Draw a qualitative graph of force \(v/s\) the distance of the charge from the centre of the ring.
(v) Calculate the time period of that S.H.M for small oscillation if \(q\) is replaced by \(- q\) and ring is fixed.
Sol. (i) Take an elementry charge dQ on the ring
Force due to this elementary charge on the point charge will be \(dF = \frac{qdQ}{4\pi\epsilon_{0}r^{2}}\)
This force can be resolved into two rectangular components one along the axis ( \(dFcos\theta\) ) and other along perpendicular to the axis ( \(dFsin\theta\) ). If we take another elementary charge diametrically opposite to dQ we observe that the vector sum of \(dFsin\theta\) for these two charge become zero. Hence vector sum of \(dFsin\theta\) for all the charges result to zero. So the resultant force on the charged particle will be only sum of \(dFcos\theta\). Therefore total force on the point charge due to

entire ring will be given by

\[F = \int_{}^{}\ dFcos\theta = \frac{1}{4\pi\varepsilon_{0}}\frac{Qx}{\mathcal{l}^{3}}\int_{}^{}\ dq = - \frac{1}{4\pi\varepsilon_{0}}\frac{Qqx}{\left( R^{2} + x^{2} \right)^{3/2}}\]

(ii) for centre \(x = 0\ \Rightarrow \text{ }F = 0\)
(iii) \(F = \frac{Qq}{4\pi\epsilon_{0}}\frac{x}{\left( R^{2} + x^{2} \right)^{3/2}}\)
\[\frac{dF}{dx} = \frac{Qq}{4\pi\epsilon_{0}}\left\lbrack \frac{\left( R^{2} + x^{2} \right)^{3/2}\frac{d}{dx}x - x\frac{d}{dx}\left( R^{2} + x^{2} \right)^{3/2}}{\left\{ \left( R^{2} + x^{2} \right)^{3/2} \right\}^{2}} \right\rbrack = \frac{Qq}{4\pi\epsilon_{0}}\left\lbrack \frac{\left( R^{2} + x^{2} \right)^{3/2}(1) - x\frac{3}{2}\left( R^{2} + x^{2} \right)^{1/2}(2x)}{\left( R^{2} + x^{2} \right)^{3}} \right\rbrack\]

\[= \left( \frac{Qq\sqrt{R^{2} + x^{2}}}{4\pi\epsilon_{0}} \right)\left( R^{2} - 2x^{2} \right)\]

For force to be maximum or minimum

\[\frac{dF}{dx} = 0 \Rightarrow R^{2} - 2x^{2} = 0 \Rightarrow x = \pm \frac{R}{\sqrt{2}}\]

(iv)

(v)

\[F = \frac{Q( - q)x}{4\pi\epsilon_{0}\left( R^{2} + x^{2} \right)^{3/2}}\]

if \(x \ll R\) then

\[F = - \frac{Qqx}{4\pi\varepsilon_{0}R^{3}}\]

Now \(\ ma = - \frac{Qqx}{4\pi\varepsilon_{0}R^{3}}\)

\[\frac{d^{2}x}{dt^{2}} = - \frac{Qq}{4\pi\varepsilon_{0}R^{3}m}x = - kx\]

\[\therefore\ k = \frac{{Qq}^{3}}{4\pi\varepsilon_{0}R^{3}\text{ }m}\]

\[T = 2\pi\sqrt{\frac{4\pi\varepsilon_{0}{mR}^{3}}{qQ}}\]

Q. 4 A small point mass \(m\) has a charge \(q\), which is constrained to move inside a narrow frictionless cylinder. At the base of the cylinder is a point mass of charge \(Q\) having the same sign as \(q\). Show that if he mass \(m\) is displaced by a small amount from its equilibrium position and released, it will exhibit simple harmonic motion with angular frequency \(\omega = \left( 2\text{ }g/y_{0} \right)^{1/2}\) where \(y_{0}\) is the equilibrium position of charge q .
Sol. In equilibrium position, gravitational force is balanced by coulombic repulsive force

\[mg = \frac{Qq}{4\pi\varepsilon_{0}y_{0}^{2}}\]

If charge q is displaced in positive y -direction, such that \(y \ll y_{0}\) from Newton's second law,

\[\begin{matrix} & \frac{Qq}{4\pi\varepsilon_{0}\left( y_{0} + y \right)^{2}} - mg = ma \\ & \frac{Qq}{4\pi\varepsilon_{0}y_{0}^{2}}\left\lbrack \frac{1}{\left( 1 + y/y_{0} \right)^{2}} \right\rbrack - mg = ma \end{matrix}\]

or \(\ mg\left\lbrack 1 - \frac{2y}{y_{0}} \right\rbrack - mg = ma\)
or \(\ a = - \frac{2gy}{y_{0}}\)

\[\frac{d^{2}y}{{dt}^{2}} + \frac{2\text{ }g}{y_{0}} = 0\]

Which is equation for SHM with \(\omega = \sqrt{\frac{2\text{ }g}{y_{0}}}\)
Q. 5 Two small identical balls lying on a horizontal plane are connected by a weightless spring. One ball (ball 2 ) is fixed at O and the other (ball 1) is free. The balls are changed identically as a result of which the spring length increases \(\eta = 2\) times Determine the change in frequency.

Sol. When the balls are uncharged, \(v_{0} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\), where k is force constant of the spring and \(m = m\) ass of the oscillating ball (ball 1 ). When charged we have in the equilibrium position of ball 1 ,

\[\begin{matrix} & \frac{1}{4\pi\varepsilon_{0}(\eta l)^{2}} = k(nl - l) = kl(\eta - 1) \\ \Rightarrow \ & l^{3} = \frac{q^{2}}{4\pi\varepsilon_{0}\eta^{2}(\eta - 1)k} \end{matrix}\]

When the ball 1 is displaced by a small distance from the equilibrium position to the right, the unbalanced force the right is given by
Resultant force to the right

\[= \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{(\eta l - x)^{2}} - k(\eta l + x - l)\]

From Newton's law, we have

\[\begin{matrix} & m\frac{{\text{ }d}^{2}x}{{dt}^{2}} = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{\eta^{2}l^{2}}\left( 1 + \frac{x}{\eta l} \right)^{- 2} - kl(\eta - 1) - kx \\ & \ = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{\eta^{2}l^{2}}\left( 1 - \frac{2x}{\eta l} \right) - kl(\eta - 1) - kx \end{matrix}\]

\[\begin{matrix} & = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{\eta^{2}l^{2}} - \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{\eta^{2}l^{2}} \cdot \frac{2x}{\eta l} - kl(\eta - 1) - kx \\ & = - \left( \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{2q^{2}}{\eta^{3}l^{3}} + k \right)x \\ & = \left( \frac{1}{4\pi\varepsilon_{0}}\frac{2q^{2}}{\eta^{3}} + k \right)x \\ & \text{ }m\frac{q^{2}x}{4\pi\varepsilon_{0}\eta^{2}(\eta - 1)k} = - \left( \frac{2(\eta - 1)}{\eta}k + k \right)x = \frac{3\eta - 2}{\eta}kx \\ & \frac{{\text{ }d}^{2}x}{{dt}^{2}} = - \frac{3\eta - 2}{\eta}\frac{k}{\text{ }m}x \\ \text{~}\text{or}\text{~}\ & \omega^{2} = \frac{3\eta - 2}{\eta}\frac{k}{\text{ }m} \\ \Rightarrow \ & v = \frac{1}{2\pi}\sqrt{\frac{3\eta - 2}{\eta}\frac{k}{\text{ }m}} \\ \text{~}\text{or}\text{~}\ & \frac{v}{v_{0}} = \sqrt{\frac{3\eta - 2}{\eta}} \end{matrix}\]

Thus the frequency is increased \(\sqrt{\frac{3\eta - 2}{\eta}}\) times.
Hence \(h = 2\) and so frequency increases \(\sqrt{2}\) times.
Q. 6 Four point charges (each having charge \(q\) ) are placed at corners of a square of side a. Find the intensity of the field at the (a) Centre of the square (b) Mid point of any side.
Sol. (a) Let position of charges be \(A,B,C\) and D and O is centre of square
Let field at O due to \(A,B,C,D\) respectively be \({\overrightarrow{E}}_{A},{\overrightarrow{E}}_{B},{\overrightarrow{E}}_{C}\) and \({\overrightarrow{E}}_{D}\). At O magnitude of field due to each point charge are same as \(OA = OB = OC = OD\) thus

\[\begin{matrix} & E_{A} = E_{B} = E_{C} = E_{D} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{{OA}^{2}} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{(a/\sqrt{2})^{2}} = \frac{1}{2\pi\varepsilon_{0}}\frac{q}{a^{2}} \\ & {\overrightarrow{E}}_{A} = \frac{1}{2\pi\varepsilon_{0}}\frac{q}{a^{2}}\left( \frac{\widehat{i}}{\sqrt{2}} + \frac{\widehat{j}}{\sqrt{2}} \right) \\ & {\overrightarrow{E}}_{B} = \frac{1}{2\pi\varepsilon_{0}}\frac{q}{a^{2}}\left( \frac{- \widehat{i}}{\sqrt{2}} + \frac{\widehat{j}}{\sqrt{2}} \right)\ C \end{matrix}\]

\[\begin{matrix} & {\overrightarrow{E}}_{C} = \frac{1}{2\pi\varepsilon_{0}}\frac{q}{a^{2}}\left( \frac{- \widehat{i}}{\sqrt{2}} - \frac{\widehat{j}}{\sqrt{2}} \right) \\ & {\overrightarrow{E}}_{D} = \frac{1}{2\pi\varepsilon_{0}}\frac{q}{a^{2}}\left( \frac{\widehat{i}}{\sqrt{2}} - \frac{\widehat{j}}{\sqrt{2}} \right) \\ & \overrightarrow{E} = {\overrightarrow{E}}_{A} + {\overrightarrow{E}}_{B} + {\overrightarrow{E}}_{C} + {\overrightarrow{E}}_{D} = 0 \end{matrix}\]

(b) Let concerned point be P

\[\begin{matrix} & \left| {\overrightarrow{E}}_{A} \right| = \left| {\overrightarrow{E}}_{B} \right| = \frac{1}{4\pi\epsilon_{0}}\frac{q}{(AP)^{2}} \\ & {\overrightarrow{E}}_{A} = \frac{q}{\pi\epsilon_{0}a^{2}}\widehat{i}\ {\overrightarrow{E}}_{B} = \frac{q}{\pi\epsilon_{0}a^{2}}( - \widehat{i}) \\ & \left| {\overrightarrow{E}}_{D} \right| = \left| {\overrightarrow{E}}_{C} \right| = \frac{1}{4\pi\epsilon_{0}}\frac{q}{(DP)^{2}} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{\left( \frac{\sqrt{5}}{2}a \right)^{2}} = \frac{q}{\pi\epsilon_{0}5a^{2}} \\ & {\overrightarrow{E}}_{C} = \frac{q}{\pi\epsilon_{0}5a^{2}}\left\lbrack \frac{- 2}{\sqrt{5}}\widehat{i} - \left( \frac{1}{\sqrt{5}} \right)\widehat{j} \right\rbrack \\ \Rightarrow \ & {\overrightarrow{E}}_{D} = \frac{q}{\pi\epsilon_{0}5a^{2}}\left\lbrack \frac{2}{\sqrt{5}}\widehat{i} - \left( \frac{1}{\sqrt{5}} \right)\widehat{j} \right\rbrack \\ \Rightarrow \ & \overrightarrow{E} = {\overrightarrow{E}}_{A} + {\overrightarrow{E}}_{B} + {\overrightarrow{E}}_{C} + {\overrightarrow{E}}_{D} = - \frac{q}{\pi\varepsilon_{0}5a^{2}\sqrt{5}}(\widehat{j}) \end{matrix}\]

Q. 7 A thin insulating rod is hinged about one of its ends; it can rotate on a smooth surface in a horizontal plane. The charge density on the rod is defined as

\[\begin{matrix} \lambda_{1} = {ax}^{2}, & 0 < x \leq \frac{L}{2} \\ \lambda_{2} = - {bx}^{n}, & \frac{\text{ }L}{2} \leq x < L \end{matrix}\]

where a and b are positive constants. An electric field \(E_{0}\), in the horizontal plane and perpendicular to the rod is switched on. Find the value of \(b\) and \(n\), if the rod has to remain stationary.

Sol. According to given charge distribution the rod is positively charge in \(0 < x \leq L/2\) and negatively charge
in \(\frac{L}{2} \leq x \leq L\). The electric force acting on the rod produce torque about hinge; for equilibrium ne torque should zero.
Torque on differential element of left half of the rod.

\[\begin{matrix} & d\tau_{1} = E_{0}\left( \lambda_{1}dx \right)x \\ & \tau_{1} = \int_{0}^{1/2}\mspace{2mu}\mspace{2mu} E_{0}\left( {ax}^{2}dx \right)x \\ & \ = \frac{E_{0}{aL}^{4}}{2^{6}} \end{matrix}\]

Torque on the right half of the rod.

\[\begin{matrix} & d\tau_{2} = - E_{0}\left( \lambda_{2}dx \right)x \\ & \tau_{2} = - \int_{L^{'}/2}^{L}\mspace{2mu}\mspace{2mu} E_{0}(bxdx)x \\ & \ = \frac{E_{0}\text{ }b}{n + 2}\frac{2^{n + 2} - 1}{2^{n + 2}}{\text{ }L}^{n + 2} \end{matrix}\]

According to condition of the problem,

\[\begin{matrix} & \left| \tau_{1} \right| = \left| \tau_{2} \right| \\ & \frac{E_{0}aL^{4}}{2^{6}} = \frac{E_{0}b}{(n + 2)}\left( \frac{2^{n + 2} - 1}{2^{n + 2}} \right)L^{n + 2} \end{matrix}\]

On comparing coefficient of L , we get \(n + 2 = 4\) or \(n = 2\) and

\[\frac{a}{2^{6}} = \frac{b}{(n + 2)}\left( \frac{2^{n + 1} - 1}{2^{n + 2}} \right)\]

or

\[b = \frac{a}{15}\]

Q. 8 In an electron ray tube, an electron enters an electric field E between the two plates with a velocity \(V_{x}\) as shown in figure and emerges from the field with a velocity V so as to strike the screen. The separation of the screen from the centre of the plate is D and length of the plate is L . If the charge of the electron is e , then the deflection Y on the screen is ( m is the mass of electron)

Sol.

\[V_{y} = \frac{eE}{m}t = \frac{eE}{m}\frac{L}{V_{x}}\]

\[\begin{array}{r} tan\theta = \frac{V_{y}}{V_{x}} = \frac{eE}{m}\frac{L}{V_{x}^{2}}\#(i) \end{array}\]

Again,

\[tan\theta = \frac{Y - Y^{'}}{D - L/2}\]

or

\[\begin{array}{r} tan\theta = \frac{Y - \frac{1}{2}\left( \frac{eE}{\text{ }m} \right)\left( \frac{L}{{\text{ }V}_{x}} \right)^{2}}{D - L/2}\#(ii) \end{array}\]

From (i) and (ii), we get

\[\frac{eE}{\text{ }m}\frac{\text{ }L}{{\text{ }V}_{x}^{2}} = \frac{Y - \frac{1}{2}\frac{eE}{\text{ }m}\frac{{\text{ }L}^{2}}{{\text{ }V}_{x}^{2}}}{D - L/2}\]

or

\[Y = \frac{eELD}{{mV}_{x}^{2}}.\]

Q. 9 An electron of mass \(m_{e}\) initially at rest moves through a certain distance in a uniform electric field in time \(t_{1}\). A proton of mass \(m_{p}\) also initially at rest takes time \(t_{2}\) to move through an equal distance in this uniform electric field. Neglecting the effect of gravity find the ratio of \(\frac{t_{2}}{t_{1}}\).
Sol. Force on a charge particle in a uniform electric field

\[F = qE\]

The acceleration imparted to the particle is

\[a = \frac{qE}{\text{ }m}\]

The distance traveled by the particle in time \(t\) is

\[d = \frac{1}{2}{at}^{2} = \frac{1}{2}\left( \frac{qE}{\text{ }m} \right)t^{2}\]

For the given problem

\[\begin{matrix} \frac{t_{p}^{2}}{{\text{ }m}_{p}} = \frac{t_{e}^{2}}{{\text{ }m}_{e}} & \frac{t_{p}^{2}}{t_{e}^{2}} = \frac{m_{p}}{{\text{ }m}_{e}} \\ \Rightarrow \ \frac{t_{p}}{t_{e}} = & \sqrt{\frac{m_{p}}{{\text{ }m}_{e}}} \end{matrix}\]

Q. 10

A particle of charge \(q\) and mass \(m\) moves rectilinearly under the action of electric field \(E = A - Bx\), where \(B\) is positive constant and \(x\) is distance from the point where particle was initially at rest then find the distance traveled by the particle before coming to rest first time and acceleration of particle at that moment
Sol.

\[\begin{matrix} & & F = qE = q(\text{ }A - Bx) \\ & & ma = q(\text{ }A - Bx) \\ & a = \frac{q}{\text{ }m}(\text{ }A - Bx) & \text{(1)} \\ & \frac{vdv}{dx} = q(\text{ }A - Bx) & \text{(1)} \\ & vdv = \frac{q}{\text{ }m}(\text{ }A - Bx)dx & \text{(1)} \\ & \ \int_{0}^{0}\mspace{2mu}\mspace{2mu} vdv = \frac{q}{\text{ }m}\int_{0}^{x}\mspace{2mu}\mspace{2mu}(\text{ }A - Bx)dx & \text{(1)} \\ & Ax - \frac{Bxx^{2}}{2} = 0 & \text{(1)} \\ & x = 0,x = \frac{2\text{ }A}{\text{ }B} & \text{(2)} \end{matrix}\]

From eq. (1) and (2)

\[\begin{matrix} & \frac{q}{m}(A - Bx) = \frac{q}{m}\left( A - B \times \frac{2A}{B} \right) \\ & \ = \frac{q}{m}(A - 2A) = \frac{- qA}{m} \end{matrix}\]

Q. 11 On a long smooth horizontal plane, over which a horizontal electric field E exists, a charged ball with mass \(m\) and charge \(q\) is dropped from a height h over the plane. Coefficient of restitution for collision between plane and ball is e . Find the ratio of maximum height attained and horizontal distance moved during the interval of nth collision to ( \(n + l\) )th collision.
Sol. Time taken to first collision drop.

\[t_{1} = \sqrt{\frac{2\text{ }h}{\text{ }g}}\]

Velocity on hitting

\[V = \sqrt{2gh}\]

Vertical velocity upward after 1st and 2nd collisions,

\[\Delta t_{1} = \frac{2{\text{ }V}_{1}}{\text{ }g} = \frac{2eV}{\text{ }g}\]

Vertical velocity upward after 2nd and 3rd collisions,

\[V_{2} = eV_{1} = e^{2}V\]

Time gap between 2 nd and 3 rd collisions,

\[\Delta t_{2} = \frac{2{\text{ }V}_{1}}{\text{ }g} = \frac{2eV}{\text{ }g}\]

Vertical velocity after nth collision,

\[V_{n} = e^{n}V\]

Maximum height attained during time gap between \(n\)th to \((n + 1)\) th collision,

\[\begin{matrix} & h = \frac{\left( e^{n}\text{ }V \right)^{2}}{2\text{ }g} \\ & \Delta t_{n} = \frac{2e^{n}\text{ }V}{\text{ }g} \end{matrix}\]

During these collisions, the ball is moving with acceleration \(\frac{qE}{m}\) horizontally.
Distance moved from nth collisions \(T_{n}\) to ( \(n + 1\) )th
Collision \(T_{n + 1}\),

\[\begin{matrix} x & \ = \frac{1}{2}a\left( T_{n + 1}^{2} - T_{n}^{2} \right) = \frac{1}{2}\left( T_{n + 1}^{2} + T_{n}^{2} \right)\left( T_{n + 1}^{2} - T_{n}^{2} \right) \\ & T_{n} = t_{1} + \left( \Delta t_{1} + \Delta t_{2} + \ldots..\Delta t_{n + 1} \right) \\ & \ = \sqrt{\frac{2h}{g}} + \frac{2V}{g}\left( e + e^{2} + \ldots.e^{n - 1} \right) \\ & \ = \sqrt{\frac{2h}{g}} + \frac{2V}{g}e\left( \frac{1 - e^{n - 1}}{1 - e} \right) \\ & T_{n + 1} = t_{1} + \left( \Delta t_{1} + \Delta t_{2} + \ldots.\Delta t_{n} \right) \\ & \ = \sqrt{\frac{2h}{g}} + \frac{2V}{g}e\left( \frac{1 - e^{n}}{1 - e} \right) \\ \Rightarrow \ & x = \frac{1}{2}\frac{eE}{m}\left\lbrack 2\sqrt{\frac{2h}{g}} + \frac{2V}{g}\frac{e}{1 - e}\left( 2 - e^{n} - e^{n - 1} \right) \right\rbrack \times \left\lbrack \frac{2V}{g}\frac{e}{1 - e}\left( e^{n - 1} - e^{n} \right) \right\rbrack \\ & \frac{h}{x} = \frac{e^{2n}V^{2}}{2g}\frac{2m}{eE\left\lbrack 2\sqrt{\frac{2h}{g}} + \frac{2V}{g}\frac{e}{1 - e}\left( 2 - e^{n} - e^{n - 1} \right) \right\rbrack} \cdot \frac{g}{2Ve^{n}} \\ & \ = \frac{mVe^{n - 1}}{2E\left\lbrack 2\sqrt{\frac{2h}{g}} + \frac{2V}{g}\frac{e}{1 - e}\left( 2 - e^{n} - e^{n - 1} \right) \right\rbrack} \\ & \ = \frac{mge^{n - 1}}{4E\left( 1 + e - e^{n} - e^{n + 1} \right)} \end{matrix}\]

Q. 12 Find the interaction force between ring of uniform charge \(Q\) and rod having uniform linear charge density \(\lambda\).

Sol.
Let us take an elementry charge on the line charge of length dx and at distance x from centre of ring. charge of elementry part will be

\[dq = \lambda dx\]

Electric field due to ring at the location of elementry charge will be

\[E = \frac{kQx}{\left( R^{2} + x^{2} \right)^{3/2}}\]

Force due to ring on elementry charge will be

\[dF = Edq = \frac{kQx\lambda dx}{\left( R^{2} + x^{2} \right)^{3/2}}\]

Therefore the total force on the rod due to ring will be

\[\begin{matrix} & F = \int_{a}^{b}\mspace{2mu}\mspace{2mu}\frac{kQx\lambda dx}{\left\lbrack R^{2} + x^{2} \right\rbrack^{3/2}} \\ & \ = kQ\lambda\left\lbrack \frac{1}{\sqrt{R^{2} + a^{2}}} - \frac{1}{\sqrt{R^{2} + b^{2}}} \right\rbrack \end{matrix}\]

Q. 13 Two point charge \(Q_{1}\) and \(Q_{2}\) lie along a line, at a distance from each other. Figure shows the potential variation along the line of charge. At which points 1,2 and 3 is the electric field zero ? What are the sings of the charges \(Q_{1}\) and \(Q_{2}\) and which of the two charges is greater in magnitude?

Sol. The electric field vector is zero at point \(3.As - \frac{dV}{dr} = E\), the negative of slop of \(V\) vs \(r\) curve represents component of electric field along r. Slope of curve is zero only at 3.
Near positive charge net potential is positive and negative near a negative charge. Thus charge \(Q_{1}\) is positive and \(Q_{2}\) negative. From the graph it can be seen that net potential due to two charge is positive in the region left of charge \(Q_{1}\) is greater than due to \(Q_{2}\). Therefore absolute value of charge \(Q_{1}\) is greater
than due to \(Q_{2}\). Secondly, the point 1 where potential due to two charge is zero, is nearer to charge \(Q_{2}\) thereby implying the \(Q_{1}\) has greater absolute value.
Q. 14 In the figure shown find the flux of electric field on curve surface. Assume that electric field is uniform

Sol. As the field is uniform ne flux on the total surface is zero.

\[\begin{matrix} \Rightarrow & \phi_{\text{curved surface}\text{~}} + \phi_{\text{Plane surface}\text{~}} = 0 \\ \Rightarrow & \phi_{\text{curved surface}\text{~}} = - \phi_{\text{Plane surface}\text{~}} = - E\pi r^{2}cos180^{\circ} = E\pi r^{2} \end{matrix}\]

Q. 15 A sphere of radius R is made of a nonconducting material that has a uniform volume charge density \(\rho\). (Assume that the material does not affect the electric field.) A spherical cavity of radius ( \(r < R\) ) is now removed from the sphere such that centre of the cavity is at position \(\overrightarrow{a}\) with respect to the centre of sphere. Show that the electric field within the cavity is uniform and is given by \(\overrightarrow{E} = \frac{\rho}{3\varepsilon_{0}}\overrightarrow{a}\)
Sol. The cavity can be considered as superposition of two charge density \(+ \rho\) and \(- \rho\) Electric field due to positive charge

\[{\overrightarrow{E}}_{+} = \frac{\rho}{3\epsilon_{0}}\overrightarrow{\mathcal{l}}\]

Electric field due to negative charge

\[\begin{matrix} & {\overrightarrow{E}}_{-} = \frac{( - \rho)}{3\epsilon_{0}}{\overrightarrow{\mathcal{l}}}^{'} \\ & \ \therefore{\overrightarrow{E}}_{\text{net}\text{~}} = {\overrightarrow{E}}_{+} + {\overrightarrow{E}}_{-} = \frac{\rho}{3\epsilon_{0}}\overrightarrow{\mathcal{l}} + \frac{( - \rho)}{3\epsilon_{0}}{\overrightarrow{\mathcal{l}}}^{'} = \frac{\rho}{3\epsilon_{0}}\left( \overrightarrow{\mathcal{l}} - {\overrightarrow{\mathcal{l}}}^{'} \right) = \frac{\rho}{3\epsilon_{0}}\overrightarrow{a} \end{matrix}\]

Q. 16 Two charge \(+ q_{1}\) and \(q_{2}\) are placed at A and B respectively. A line of force emanates from \(q_{1}\) at angle \(\alpha\) with the line \(AB\). At what angle will it terminate at \(- q_{2}\) ?

Sol. It is the property of the line of force that their number within a tube remains unchanged and the number of line of force is equal to the charge. The line of force emanating from \(q_{1}\) spreads out equally in all directions. Hence lines of force per unit solid angle are \(\frac{q_{1}}{4\pi}\) and the number of lines through cone of halfangle \(\alpha\) is \(\frac{q_{1}}{4\pi} \cdot 2\pi(1 - cos\alpha)\) because solid angle of a cone is \(2\pi(1 - cos\alpha)\).

Similarly the number of line of force terminating on \(- q_{2}\) at \(\beta\) is

\[\frac{q_{2}}{4\pi} \cdot 2\pi(1 - cos\beta)\]

By the property of lines of force.

\[\frac{q_{1}}{4\pi} \cdot 2\pi(1 - cos\alpha) = \frac{q_{2}}{4\pi} \cdot 2\pi(1 - cos\beta)\]

\[\begin{matrix} \Rightarrow & \frac{q_{1}}{2} \cdot 2\sin^{2}\frac{\alpha}{2} = \frac{q_{2}}{2} \cdot 2\sin^{2}\frac{\beta}{2} \\ \Rightarrow & sin\frac{\beta}{2} = 2sin\frac{\alpha}{2}\sqrt{\frac{q_{1}}{q_{2}}} \\ \Rightarrow & \beta = 2\sin^{- 1}\left( sin\frac{\alpha}{2}\sqrt{\frac{q_{1}}{q_{2}}} \right) \end{matrix}\]

Q. 17 Two particles of mass \(m\) and \(2m\) carry a charge \(q\) each. Initially the heavier particle is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first from a distance d with speed \(u\). Find the closest distance of approach.
Sol. As the mass 2 m is not fixed, it will also move away from m due to repulsion. The distance between the particles is minimum when their relative velocity is zero i.e., when they have equal velocities.

Hence at closest approach, \(\ v_{1} = v_{2}\)
By conservation of momentum

\[\begin{matrix} & mu = {mv}_{1} + 2{mv}_{2} \\ & v_{2} = v_{1} = u/3 \end{matrix}\]

By conservation of energy
Loss in \(KE =\) gain in PE

\[\begin{matrix} & \frac{1}{2}{mu}^{2} - \left( \frac{1}{2}{mv}_{1}^{2} + \frac{1}{2}2{mv}_{2}^{2} \right) = \frac{q^{2}}{4\pi\varepsilon_{0}}\left( \frac{1}{x} - \frac{1}{\text{ }d} \right) \\ & \frac{1}{2}{mu}^{2} - \frac{1}{2}\text{ }m\frac{u^{2}}{9}(1 + 2) = \frac{q^{2}}{4\pi\varepsilon_{0}}\left( \frac{1}{x} - \frac{1}{\text{ }d} \right) \\ & \frac{1}{3}{mu}^{2} = \frac{q^{2}}{4\pi\varepsilon_{0}}\left( \frac{1}{x} - \frac{1}{\text{ }d} \right) \\ & \frac{1}{x} = \frac{1}{\text{ }d} + \frac{4\pi\varepsilon_{0}{mu}^{2}}{3q^{2}} \\ & x = \frac{3q^{2}\text{ }d}{3q^{2} + 4\pi\varepsilon_{0}{mu}^{2}\text{ }d}. \end{matrix}\]

Q. 18 A charged particle having a charge \(q\) moves along the \(x\)-axis with a constant velocity \(v_{0}\). Another particle \(B\) with charge \(q\) and mass \(m\) is lying on thy \(y\)-axis at \(y = a\). The particle \(B\) is constrained to move along the \(y\)-axis. While the particle \(A\) moves along the \(x\)-axis.
Assuming that the velocity \(v_{0}\) is very large, find the impulse imparted to \(B\) along, the \(y\)-axis as the particle A moves from \(- \infty\) to \(\infty\), assuming that the motion of particle \(B\) is negligible.

Sol. First we will determine the angular speed a A relative to B at angular position \(\theta\) shown in Fig. For particle A,

\[\begin{array}{r} \frac{dx}{dt} = \frac{d}{dt}(attan\theta) = v_{0}\#(i) \end{array}\]

or \(\ a\sin^{2}\theta\frac{d\theta}{dt} = v_{0}\)
or \(\ \left( \frac{d\theta}{dt} \right) = \frac{v_{0}}{a}\cos^{2}\theta\)

As particle A is moving very fast, we can assume that when it crosses along the x -axis, the y component of the force on B due to A is

\[\begin{array}{r} F_{y} = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{a^{2}\sec^{2}\theta} \times cos\theta\#(iii) \end{array}\]

Impulse on

\[\begin{matrix} & B = \int_{}^{}\ F_{y}dt = \int_{}^{}\ \frac{F_{y}\text{ }d\theta}{(\text{ }d\theta/dt)} \\ & \ = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{a^{2}}\frac{a}{v_{0}}\int_{}^{}\ \frac{\cos^{3}}{\cos^{2}}\text{ }d\theta \\ & \ = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{{av}_{0}}\int_{\theta = - /2}^{\pi/2}\mspace{2mu}\mspace{2mu} cos\theta\text{ }d\theta \end{matrix}\]

The impulse delivered to B ,

\[\begin{array}{r} = \Delta P_{y} = \frac{1}{4\pi\varepsilon_{0}}\frac{2q^{2}}{{av}_{0}} = \frac{q^{2}}{2\pi\varepsilon_{0}\left( {av}_{0} \right)}\#(iv) \end{array}\]

Q. 19 Three point charges \(q,2q\) and 8 q are to be placed on a 9 cm long straight line. Find the positions where the charges should be placed such that the potential energy of this system is minimum. In this situation, what is the electric field at the charge \(q\) due to the other two charges ?
Sol. The maximum contribution may come from the charge \(8q\) forming pairs with others. To reduce its effect, it should be placed at a corner and the smallest charge \(q\) in the middle. This arrangement shown in figure ensures that the charges in the strongest pair \(2q,8q\) are at the largest separation.
The potential energy is

\[U = \frac{q^{2}}{4\pi\varepsilon_{0}}\left\lbrack \frac{2}{x} + \frac{16}{9\text{ }cm} + \frac{8}{9\text{ }cm - x} \right\rbrack.\]

This will be minimum if

\[A = \frac{2}{x} + \frac{8}{9\text{ }cm - x}\text{~}\text{is minimum.}\text{~}\]

For this, \(\frac{dA}{dx} = - \frac{2}{x^{2}} + \frac{8}{(9\text{ }cm - x)^{2}} = 0\)
or, \(9\text{ }cm - x = 2x\) or, \(x = 3\text{ }cm\)
The electric field at the position of charge \(q\) is

\[\begin{matrix} & \frac{q}{4\pi\varepsilon_{0}}\left( \frac{2}{x^{2}} - \frac{8}{(9\text{ }cm - x)^{2}} \right) \\ & \ = 0 \end{matrix}\]

Q. 20 Two points charge q and -2 q are placed at a distance 6 a apart. Find the locus of the point in the plane of charges where the field potential is zero.
Sol. Let us take the charge on X -axis;
q at \(A(0,0)\) and -2 q at \(B(6a,0)\)
Potential at a point \(P(x,y)\) is

\[\begin{matrix} & V = \frac{q}{4\pi\varepsilon_{0}\sqrt{x^{2} + y^{2}}} + \frac{- 2q}{4\pi\varepsilon_{0}\sqrt{(x - 6a)^{2} + y^{2}}} \\ & V = 0 \\ \Rightarrow \ & \frac{q^{2}}{x^{2} + y^{2}} = \frac{4q^{2}}{(x - 6a)^{2} + y^{2}} \\ \Rightarrow \ & \text{~}\text{the locus is}\text{~}(x - 6a)^{2} = 4x^{2} + 3y^{2} \\ \Rightarrow \ & 3x^{2} + 3y^{2} + 2(6a)x = 36a^{2} \\ & x^{2} + y^{2} + 4ax = 12a^{2} \\ & (x + 2a)^{2} + y^{2} = 16a^{2} \end{matrix}\]

A circle with centre (-2a, 0 ) and radius 4a.
Q. 21 The \(arcAB\) with the centre C and the infinitely long wire having linear charge density \(\lambda\) are lying in the same plane. Find the minimum amount of work to be expended to move a point charge \(q_{0}\) from point A to \(B\) through a circular path \(AB\) of radius \(a\) is equal to :

Sol. \(E = \frac{\lambda}{2\pi\epsilon_{0}x}\)

\[\begin{matrix} & \int_{V_{A}}^{V_{B}}\mspace{2mu}\mspace{2mu} dV = - Edx = - \frac{\lambda}{2\pi\varepsilon_{0}}\int_{3a}^{2a}\mspace{2mu}\mspace{2mu}\frac{dx}{x} \\ \Rightarrow \ & {\text{ }V}_{B} - V_{A} = \frac{\lambda}{2\pi\epsilon_{0}}ln\left( \frac{3}{2} \right) \end{matrix}\]

work done by agent \(= \frac{q_{0}\lambda}{2\pi\varepsilon_{0}}In\frac{3}{2}\).
Q. 22 A circular ring of radius R with uniform positive charge density \(\lambda\) per unit length is located in the \(Y - Z\) plane with its centre at the origin \(O\). A particle of mass \(m\) and positive charge \(q\) is projected from the point \(P(R\sqrt{3},0,0)\) on the positive x -axis directly towards O , whit initial velocity v . Find the smallest (non-zero) value of the speed \(v\) such that the particle does not returen to \(P\).

Sol. The situation is shown in figure. Total potential at the centre of a ring is given by

\[\begin{matrix} & V = \frac{\lambda R}{2\varepsilon_{0}\sqrt{\left( a^{2} + R^{2} \right)}} \\ & \ = \frac{\lambda R}{2\varepsilon_{0}\sqrt{\left\lbrack (\sqrt{3}R)^{2} + R^{2} \right)}} = \frac{\lambda}{4\varepsilon_{0}} \end{matrix}\]

Potential energy at \(P = \lambda q/4\varepsilon_{0}\)
Energy of particle at \(P = \frac{1}{2}{mv}^{2}\)
\(\therefore\ \) Total energy at \(P = \frac{\lambda q}{4\varepsilon_{0}} + \frac{1}{2}{mv}^{2}\)
The potential energy at centre \(= \left( \lambda q/2\varepsilon_{0} \right)\)
The particle will not return to \(P\), when

\[\begin{matrix} & \frac{\lambda q}{4\varepsilon_{0}} + \frac{1}{2}mv^{2} = \frac{\lambda q}{2\varepsilon_{0}} \\ & \frac{1}{2}mv^{2} = \frac{\lambda q}{2\varepsilon_{0}} - \frac{\lambda q}{4\varepsilon_{0}} = \frac{\lambda q}{4\varepsilon_{0}} \\ & v^{2} = \frac{\lambda q}{2\varepsilon_{0}m} \\ & v = \sqrt{\left( \frac{\lambda q}{2\varepsilon_{0}m} \right)} \end{matrix}\]

Q. 23 Asmall positively charge ball of mass \(m\) is suspended by an insulating thread of negligible mass. Another positively charged small ball is moved very slowly from a large distance until it is in the original position of the first ball. As a result, the first ball rises by \(h\). How much work has been done?

(b)

Sol. Using the notation of figure the equilibrium condition for the first ball is

\[s\frac{mg}{\text{ }F} = \frac{l}{x}\]

Where \(F = kqQ/x^{2}\) is the Coulomb force acting on the first ball and x is the distance between the balls carrying charges q and Q .
It is clear the triangles ABD and CAE are similar and that consequently

\[\frac{x}{2}:l = h:x\]

From the three equations above we can calculate the separation of the charge and the electrostatics energy of the system:

\[x = k\frac{qQ}{2mgh}\]

and

\[\begin{matrix} & E_{\text{electro}\text{~}} = k\frac{qQ}{x} \\ & \ = 2mgh. \end{matrix}\]

The work done is the sum of the changes in electrical and gravitational potential energy.

\[\begin{matrix} & W = 2mgh + mgh \\ & \ = 3mgh \end{matrix}\]

Note that the work done does not depend on either the magnitudes of the charges or the length of the thread.
Q. 24 A charge Q is uniformly distributed over a spherical volume of radius R. Obtain an expression for the energy of the system.
Sol. In this case, the electric field exists from centre of the sphere to infinity. Potential energy is stored in electric field with energy density.

\[u = \frac{1}{2}\varepsilon_{0}E^{2}\text{~}\text{(energy/volume)}\text{~}\]

(i) Energy stored within the sphere \(\left( V_{1} \right)\)

Energy field at a distance r is ( \(r \leq R\) )

\[\begin{matrix} & E = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{Q}{R^{3}} \cdot r \\ \therefore & u = \frac{1}{2}\varepsilon_{0}E^{2} \\ & u = \frac{\varepsilon_{0}}{2}\left\{ \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{Q}{R^{3}}r \right\}^{2} \end{matrix}\]

Volume of element, \(dV = \left( 4\pi r^{2} \right)dr\)
\(\therefore\ \) Energy stored in this volume, \(dU = u(dV)\)

\[\begin{matrix} & dU = \left( 4\pi r^{2}dr \right)\frac{\varepsilon_{0}}{2}\left\{ \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{Q}{R^{3}}r \right\}^{2} \\ & dU = \frac{1}{8\pi\varepsilon_{0}} \cdot \frac{Q^{2}}{R^{6}} \cdot r^{4}dr \\ & U_{1} = \int_{0}^{R}\mspace{2mu}\mspace{2mu} dU = \frac{1}{8\pi\varepsilon_{0}}\frac{Q^{2}}{R^{6}}\int_{0}^{R}\mspace{2mu}\mspace{2mu} f^{4}dr \\ & = \frac{Q^{2}}{40\pi\varepsilon_{0}R^{6}}\left\lbrack r^{5} \right\rbrack_{0}^{R} \\ U_{1} = \frac{1}{40\pi\varepsilon_{0}} \cdot \frac{Q^{2}}{R} & \text{(1)} \end{matrix}\]

(ii) Energy stored outside the sphere \(\left( U_{2} \right)\)

Electric field at a distance \(r\) is ( \(r \geq R\) )

\[\begin{matrix} & & E = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{Q}{r^{2}} \\ & & u = \frac{1}{2}\varepsilon_{0}E^{2} \\ & & u = \frac{\varepsilon_{0}}{2}\left\{ \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{Q}{r^{2}} \right\}^{2} \\ & & dV = \left( 4\pi r^{2}dr \right) \\ & & dU = udV = \left( 4\pi r^{2}dr \right)\left\lbrack \frac{\varepsilon_{0}}{2}\left( \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{Q}{r^{2}} \right)^{2} \right\rbrack \\ & & dU = \frac{Q^{2}}{8\pi\varepsilon_{0}} \cdot \frac{dr}{r^{2}} \\ & & U_{2} = \int_{R}^{\infty}\mspace{2mu}\mspace{2mu} dV = \frac{Q^{2}}{8\pi\varepsilon_{0}}\int_{R}^{\infty}\mspace{2mu}\mspace{2mu}\frac{dr}{r^{2}} \\ & U_{2} = \frac{Q^{2}}{8\pi\varepsilon_{0}R} & \text{(2)} \end{matrix}\]

Therefore, total energy of the system is

\[\begin{matrix} & U = U_{1} + U_{2} = \frac{Q^{2}}{40\pi\varepsilon_{0}R} + \frac{Q^{2}}{8\pi\varepsilon_{0}R} \\ & U = \frac{3}{20}\frac{Q^{2}}{\pi\varepsilon_{0}R}. \end{matrix}\]

Q. 25 A dipole of mass \(m\) is placed in front of a line charge at a separation \(x\) (as shown in figure). Find (i) interaction energy between point charge and dipole (ii) force acting on the dipole due to line charge(iii) what will be its velocity if it reaches at separation \(x_{0}\)
Sol.
(i) \(U = - \overrightarrow{p} \cdot \overrightarrow{E} = - pEcos0^{\circ} = - p\frac{\lambda}{2\pi\epsilon_{0}x}(1) = - \frac{\lambda p}{2\pi\epsilon_{0}x}\)
(ii) \(F = - \frac{dU}{dx} = - \frac{\lambda p}{2\pi\epsilon_{0}x^{2}} \Rightarrow \overrightarrow{F}\frac{\lambda p}{2\pi\epsilon_{0}x^{2}}\) directed along-ve \(x\) axis
(iii) using conservation of energy
\[{K_{f} + U_{f} = K_{i} + U_{i} }{\frac{1}{2}mv^{2} - \frac{\lambda p}{2\pi\epsilon_{0}x_{0}} = 0 - \frac{\lambda p}{2\pi\epsilon_{0}x} }\]

\[{\frac{1}{2}mv^{2} = \frac{\lambda p}{2\pi\epsilon_{0}}\left( \frac{1}{x_{0}} - \frac{1}{x} \right) }{v = \sqrt{\frac{2\lambda p}{2\pi\epsilon_{0}m}\left( \frac{1}{x_{0}} - \frac{1}{x} \right)} }\]Q. 26

Figure shows three concentric thin spherical shells \(A,B\) and C of radii \(a,b\) and c repectively. The shells \(A\) and \(C\) are given charges \(q\) and \(- q\) respectively and the shell \(B\) is earthed. Find the charges appearing on the surface of B and C .

(a)

(b)

Sol. As shown in the previous worked out exmple. The inner surface of \(B\) must have a charge \(- q\) from the Gauss's law. Suppose, the outer surface of B has a charge \(q^{'}\). The inner surface of C must have a charge \(- q^{'}\) from the Gauss's law. As the net charge on C must be -q , its outer surface should have a charge q ' -q . The charge distribution is shown in figure.
The potential at \(B\) due to the charge \(q\) on \(A\)

\[= \frac{q}{4\pi\varepsilon_{0}\text{ }b},\]

due to the charge \(- q\) on the inner surface of \(B\)

\[= \frac{- q}{4\pi\varepsilon_{0}\text{ }b}\]

due to the charge \(q^{'}\) on the outer surface of \(B\)

\[= \frac{q^{'}}{4\pi\varepsilon_{0}\text{ }b}\]

due to the charge \(- q^{'}\) on the outer surface of C

\[= \frac{- q^{'}}{4\pi\varepsilon_{0}c}\]

and due to the charge \(q^{'} - q\) on the outer surface of C

\[= \frac{q^{'} - q}{4\pi\varepsilon_{0}c}\]

The net potential is

\[V_{B} = \frac{q^{'}}{4\pi\varepsilon_{0}\text{ }b} - \frac{q}{4\pi\varepsilon_{0}c}\]

This should be zero as the shell \(B\) is eartdhed. Thus,

\[q^{'} = \frac{b}{c}q.\]

The charges on various surface are as shown in figure

Q27 Figure shows three plates \(A,B\) and C each of area S in which A and C are joined by a conducting wire. Plate \(A\) and \(B\) are given charge \(Q\) and \(3Q\) respectively. Find the final charge distribution of each side of each conductor.

Sol. Middle plate is isolated. Hence sum of charge on left and right side will be 3Q. But left+right plate forms isolated system and using conservation of charge for this system.
\[y + ( - x) + \{ - (3Q - x)\} + z = 0 \]but \(y = z\)
...........(ii) \(\lbrack\because E = 0\) inside conductor \(\rbrack\)

Also left and right plate have same potential. Hence potential difference between A and C will be zero. i.e,
\[{\int_{A}^{c}\mspace{2mu}\overrightarrow{E} \cdot d\overrightarrow{r} = 0 }{\Rightarrow \int_{A}^{B}\mspace{2mu}\overrightarrow{E} \cdot d\overrightarrow{r} + \int_{B}^{C}\mspace{2mu}\overrightarrow{E} \cdot d\overrightarrow{r} = 0 }{\Rightarrow - E_{1}\text{ }d + E_{2}(2\text{ }d) = 0}\]

\[\begin{array}{r} \Rightarrow - \left\{ \frac{x}{2\epsilon_{0}\text{ }S} \times 2 \right\} d + \left\{ \frac{3Q - x}{2\epsilon_{0}\text{ }S} \times 2 \right\}(2\text{ }d) = 0\#(iii) \end{array}\]

After solving (i), (ii) and (iii) we get
\[x = 2Q\ y = z = 2Q \]