ELECTRIC FLUX AND GAUSS'S THEOREM -

Solid Angle:-

An angle formed at a point due to an area is known as a solid angle. A solid angle is a three-dimensional angle subtended by a spherical section at its centre of curvature. Its S.I. unit is steradian. It is a dimensionless quantity.

From Fig, solid angle of area dA at point O is

$\dpi{120} \fn_cm d\Omega= \frac{dA}{r^2}$

The solid angle subtended at the centre by the complete surface of the sphere is

$\dpi{120} \fn_cm \Omega=\frac{4\pi r^2}{r^2}=4\pi\;\;steradian$

From above fig

solid angle $\dpi{120} \fn_cm (\Omega)=\frac{A_1}{r_1^2}=\frac{A_2}{r_2^2}$

Relation between the half angle of the cone and the solid angle at the vertex:-

Area of the shaded portion is $\dpi{120} \fn_cm dA=(2\pi R\sin\phi)Rd\phi$

∴ Total area of spherical section is

$\dpi{120} \fn_cm A=\int dA=2\pi R^2\int_{0}^{\theta}\sin\phi d\phi$

$\dpi{120} \fn_cm A=2\pi R^2\left [ -\cos\phi \right ]_{0}^{\theta}$

$\dpi{120} \fn_cm A=2\pi R^2(-\cos\theta+ \cos 0^0)$

$\dpi{120} \fn_cm A=2\pi R^2(1-\cos\theta)$

$\dpi{120} \fn_cm \therefore$ solid angle is

$\dpi{120} \fn_cm \Omega=\frac{A}{R^2}$

$\dpi{120} \fn_cm \Omega=\frac{2\pi R^2(1-\cos\theta)}{R^2}$

$\dpi{120} \fn_cm \left [ \Omega=2\pi (1-\cos\theta) \right ]$

Solid angle subtended by hemisphere, cone or pyramid

Same case in all (θ=90°)

$\dpi{120} \fn_cm \Omega=2\pi (1-\cos 90^0)=2\pi\;\;steradian$

For any closed surface

$\dpi{120} \fn_cm \Omega=2\pi (1-\cos 180^0)=4\pi\;\;steradian$

${\color{Red} \textbf{NOTE:-}}$ Every closed surface subtended a solid angle of 4π at every interior point.

Show that the 1/r² dependence of the electric field of a point charge is consistent with the concept of the electric field lines

It is clear, that the number of lines originating from point charge q in a given solid angle ΔΩ is constant. Let n be the number of lines passing through S1 and S2.

We know that electric field strength at a point is proportional to the number of lines passing through normally per unit area.

i.e. electric field at the point $\dpi{120} \fn_cm P_1$ is

$\dpi{120} \fn_cm E_1\propto \frac{n}{r_1^2 \Delta \Omega}-----(1)$

and the electric field at the point $\dpi{120} \fn_cm P_2$ is

$\dpi{120} \fn_cm E_2\propto \frac{n}{r_2 ^2 \Delta \Omega}-----(2)$

From (1) and (2)

$\dpi{120} \fn_cm \frac{E_1}{E_2}=\frac{n}{r_1^2 \Delta\Omega}.\frac{r_2^2 \Delta\Omega}{n}$

$\dpi{120} \fn_cm \frac{E_1}{E_2}=\frac{r_2^2}{r_1^2}$

$\dpi{120} \fn_cm i.e. \left [ E\propto\frac{1}{r^2} \right ]$

Area Vector:-

Area is a scalar quantity but in some situations, it has to be treated as a vector quantity. ( Area is considered to be a vector when it comes along with other vectors.)

Here area should be flat

Thus area vector $\dpi{120} \fn_cm d\vec{A}$ is a vector whose magnitude is the area $\dpi{120} \fn_cm dA$ and its direction is along outward drawn normal to the area vector.

Electric Flux:-

(It refers to some kind of flow)

The electric flux through a given area in an electric field represents the total number of electric field lines crossing the area. It is denoted by $\dpi{120} \fn_cm \Phi$ .

Electric flux through an area is defined as the dot product of electric field strength $\dpi{120} \fn_cm \vec{E}$ and area vector $\dpi{120} \fn_cm d\vec{A}$

$\dpi{120} \fn_cm i.e. \left [ \Phi=\vec{E}.d\vec{A} \right ]$

$\dpi{120} \fn_cm \left [ \Phi=E dA\cos\theta \right ]$ where $\dpi{120} \fn_cm \theta$ is the angle between the Electric field and area vector

It is a scalar quantity and its S.I unit is Nm²/C or V-m. Its dimension is $\dpi{120} \fn_cm [ML^3T^{-3}A^{-1}]$

${\color{Red} \textbf{Special Case}}$

If $\dpi{120} \fn_cm \theta=0^0,\;\;\; \Phi=E A\cos 0^0=EA\;\;\;(+ve)$
if $\dpi{120} \fn_cm \theta=90^0,\;\;\; \Phi=E A\cos 90^0=0$
if $\dpi{120} \fn_cm \theta=180^0,\;\;\; \Phi=E A\cos 180^0=-EA\;\;\;(-ve)$

${\color{Red} \textbf{NOTE:-}}$

It is proportional to the number of field lines through an area.
It may be +ve, -ve or zero
This formula is valid only when the electric field is constant through the surface and the surface must be flat. Otherwise, use the integration method. $\dpi{120} \fn_cm \left [ \Phi=\int \vec{E}.d\vec{A} \right ]$

Exercise

(1) If $\dpi{120} \fn_cm \vec{E}=6\hat{i}+3\hat{j}+4\hat{k},$ calculate the electric flux through the surface of area 20 units in Y-Z plane. ( 120 unit)

(2) A circular plane of radius 10 cm is placed in a uniform electric field of $\dpi{120} \fn_cm 5\times 10^5 N/C,$ making an angle $\dpi{120} \fn_cm 60^0$ with the field. Calculate electric flux through the sheet. $\dpi{120} \fn_cm \left ( 1.36 \times 10^4 Nm^2/C \right )$

(3) The electric field in a certain region of space is $\dpi{120} \fn_cm (5\hat{i}+4\hat{j}-4\hat{k})\times 10^5 N/C$ . Calculate electric flux due to this field over an area of $\dpi{120} \fn_cm (2\hat{i}-\hat{j})\times 10^{-2}m^2$ . $\dpi{120} \fn_cm (6\times 10^3 \; Nm^2/C)$

(4) A closed cylinder is placed in a uniform electric field with its axis parallel to the field. Show that the total electric flux through the cylinder is zero.

(5) Consider a uniform electric field $\dpi{120} \fn_cm \vec{E}=3\times 10^3\; \hat{i}\; N/C$ . Calculate the flux of this field through the square surface of area $\dpi{120} \fn_cm 10 cm^2$ when

(a) Its plane is parallel to the Y-Z plane (30 Nm²/C)
(b) Its plane is parallel to the X-Y plane and X-Z plane (0.0)
(c) The normal to its plane makes a 60° angle with the x-axis. (15 Nm²/C)

(6) Find the flux of the field through the hemisphere. The magnitude of the field at the surface is E. (2πER²)

(7) Find the flux of a uniform electric field through the curved surface.(πER²)

(8) A cube is placed in a non-uniform electric field $\dpi{120} \fn_cm \vec{E}=3x\hat{i}\; V/m$ as shown. Find the flux through surfaces 1 and 2. Also, find flux through the remaining four surfaces. (-48 Vm, 240 Vm, 0,0,0,0)

(9) Find the flux through the given circle as shown in the figure. $\dpi{120} \fn_cm \left ( \frac{q}{\epsilon_0}\left \{ 1-\frac{1}{\sqrt{1+\left ( \frac{R}{l} \right )^2}} \right \} \right )$

(10) Find the flux of the electric field through each of the five surfaces of the inclined plane as shown in fig. (Eab sinθ, -Eab sinθ, 0,0,0,0)

(11) The cube has sides of length L=10 cm. The electric field is uniform, has a magnitude $\dpi{120} \fn_cm E=4\times10^3 N/C,$ and is parallel to the x-y plane at an angle $\dpi{120} \fn_cm 37^0$ measured from the +ve x-axis towards the y-axis. What is the electric flux through each of the six cube faces?

(12) Calculate the flux of the electric field entering the given shape as shown in the figure. (2ERh, ER(2h cosθ+πR sinθ), ERh)

(13) A uniform electric field $\dpi{120} \fn_cm a\hat{i}+b\hat{j}$ intersects a surface of area A. What is the flux through this area if the surface lies (a) in the y-z plane (b) in the x-z plane and (c) in the x-y plane? (Aa, Ab, 0)

Gauss’s Law in Electrostatic:-

( Alternative to Coulomb’s law used in symmetry\ic properties)

It states that the total electric flux passing through a closed surface is equal to $\dpi{120} \fn_cm \frac{1}{\epsilon_0}$ times the net charge enclosed by the closed surface.

i.e. $\dpi{120} \fn_cm \mathbf{\left [ \Phi=\frac{Q_{inside}}{\epsilon_0} \right ]}$

Also $\dpi{120} \fn_cm \Phi=\oint \vec{E}.d\vec{A}$

i.e. we can write

$\dpi{120} \fn_cm \left [ \oint \vec{E}.d\vec{A}=\frac{Q_{inside}}{\epsilon_0} \right ]$

${\color{Red} \textbf{NOTE:-}}$

(1) For other medium $\dpi{120} \fn_cm \left [ \Phi=\frac{Q_{inside}}{\epsilon_m}=\frac{Q_{inside}}{K \epsilon_0} \right ]\;\;\;\;\; \epsilon_r=K=\frac{\epsilon_m}{\epsilon_0}$

(2) From the above relation $\dpi{120} \fn_cm Q_{inside}$ is only charge taken inside (Placed anywhere) But $\dpi{120} \fn_cm \vec{E}$ is electric field due to both charges inside and outside.

(3) Gauss’s Law is true for any closed surface, no matter what its shape and size.

(4) If no net flux passes through any closed surface E may not be zero, but the flux of E must be zero.

(5) If we change the configuration by displacing the charges inside the Gaussian surface, the electric field at any point may change leaving the net flux passing through the Gaussian surface unchanged.

(6) However, any change in the configuration of the charges outside a Gaussian surface may change E at a point, but the net flu remains the same as external charges contribute nothing to the total flux.

Gaussian Surface:-

A hypothetical closed surface chosen to calculate the surface integral of the electric field is called a Gaussian surface. Generally on the Gaussian surface magnitude of E must be the same otherwise E⊥A.

Exercise

(1) A point charge of is at the centre of a cube of 1m side. What is the flux for the surface?

(2) Calculate the number of electric line of force originating from a charge of 1C.

(3) A +ve charge of is placed at the centre of a hollow sphere of radius. Calculate the flux density through the surface of the sphere.

(4) Calculate the electric flux through each of the six faces of a closed cube of length l, if charge q is placed

(a) At its centre and

(b) At one of its vertices (closed to vertices)

(5) Determine the flux through the hemispherical surface.

(6) If electric field strength depends only on the and coordinate as, where a is constant. Find the flux of the vector through a sphere of radius R with its centre at the origin of the coordinate. Using the above result, also calculate the total charge enclosed by the sphere.

Applications of Gauss’s Law:-

Proof of Gauss’s Theorem

Electric field at any point on S is

$\dpi{120} \fn_cm E=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$

∴ flux through a small area dA on this surface is

$\dpi{120} \fn_cm d\Phi=\vec{E}.d\vec{A}=EdA\cos0^0=EdA$

∴ Total flux through the Gaussian surface is

$\dpi{120} \fn_cm \Phi=\int EdA=E\int dA=E4\pi r^2$

$\dpi{120} \fn_cm \Phi=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}.4\pi r^2$

$\dpi{120} \fn_cm \mathbf{\left [ \Phi=\frac{q}{\epsilon_0} \right ]}$

This proof of Gauss’s theorem

Coulomb’s law from Gauss’s theorem

Consider an isolated +ve point charge $\dpi{120} \fn_cm q_1$ . We select a spherical surface S of radius r centred at charge $\dpi{120} \fn_cm q_1$ as the Gaussian surface.

Take a small area dA at point P on the Gaussian surface. By symmetry, let E be the magnitude of the electric field at all points on S.

From Fig, flux through this small area is

$\dpi{120} \fn_cm d\Phi=\vec{E}.d\vec{A}$

∴ Total flux through closed surface S is given by

$\dpi{120} \fn_cm \Phi=\int EdA=E\int dA\;\;\;\;\;\;\;(E\;is\;constant\;over\;surface)$

$\dpi{120} \fn_cm \Phi=E.4\pi r^2-----(1)$

From Gauss’s theorem

$\dpi{120} \fn_cm \Phi=\frac{q_1}{\epsilon_0}-----(2)$

From (1) and (2)

$\dpi{120} \fn_cm E.4\pi r^2=\frac{q_1}{\epsilon_0}$

$\dpi{120} \fn_cm \therefore \left [ E=\frac{1}{4\pi \epsilon_0}.\frac{q_1}{r^2} \right ]$

If a point charge $\dpi{120} \fn_cm q_2$ is placed on the surface at point P. Then the force on $\dpi{120} \fn_cm q_2$ is given by

$\dpi{120} \fn_cm F=q_2 E$

$\dpi{120} \fn_cm \mathbf{\left [ F=\frac{1}{4\pi \epsilon_0}.\frac{q_1q_2}{r^2} \right ]}$

This proves the Coulomb’s Law

Electric field due to an infinitely long straight charged wire.

Consider a thin infinitely long straight wire having a uniform linear charge density $\dpi{120} \fn_cm \lambda$ .

We have to calculate the electric field at point P which is perpendicular distance r from the rod.

Draw a Gaussian surface S, like a closed circular cylinder of radius r and arbitrary length l coaxial with the rod as shown in Fig.

The flux through the top and bottom flat surface surface is zero because $\dpi{120} \fn_cm \vec{E}\perp \vec{A}$

By symmetry electric field $\dpi{120} \fn_cm \vec{E}$ is same at every point on the curved surface.

$\dpi{120} \fn_cm \therefore$ The flux through a small area of dA on curved surface is

$\dpi{120} \fn_cm d\Phi=\vec{E}.d\vec{A}=E dA\cos0^0=EdA$

$\dpi{120} \fn_cm \therefore$ The total flux through the cylinder is

$\dpi{120} \fn_cm \Phi=\int E.dA=E\int dA$

$\dpi{120} \fn_cm \Phi=E 2\pi r l-----(1)$

From Gauss’s theorem, net flux through Gaussian surface S is

$\dpi{120} \fn_cm \Phi=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm \Phi=\frac{\lambda l}{\epsilon_0}-----(2)$

From (1) and (2)

$\dpi{120} \fn_cm E.2\pi rl=\frac{\lambda l}{\epsilon_0}$

$\dpi{120} \fn_cm \therefore \mathbf{\left [ E=\frac{1}{2\pi \epsilon_0}\frac{\lambda}{r} \right ]}$

Electric field due to uniformly charged spherical shell or solid spherical conductor

Consider a thin spherical shell of radius R with uniform charge distribution Q. We have to calculate the electric field at point P which is distance r from the centre of the shell.

From symmetry, the magnitude of the electric field is same at all points equidistance from the centre of the shell

$\dpi{120} \fn_cm {\color{Red} \textbf{Case I :-}}$ When point P lies outside the shell

Draw a Gaussian surface as shown in the figure.

The flux through a small area dA at point P on the Gaussian surface is

$\dpi{120} \fn_cm d\Phi=\vec{E}.d\vec{A}=E dA\cos0^0=EdA$

∴ Flux through the total Gaussian surface is

$\dpi{120} \fn_cm \Phi=\int E.dA=E\int dA$

$\dpi{120} \fn_cm \Phi=E 4\pi r^2-----(1)$

From Gauss’s theorem, total flux through the Gaussian surface is

$\dpi{120} \fn_cm \Phi=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm \Phi=\frac{Q}{\epsilon_0}-----(2)$

From (1) and (2)

$\dpi{120} \fn_cm E.4\pi r^2=\frac{Q}{\epsilon_0}$

$\dpi{120} \fn_cm \therefore \mathbf{\left [ E=\frac{1}{4\pi \epsilon_0}.\frac{Q}{r^2} \right ]}$

i.e. Formula is such that, the total charge is concentrated at the centre of shell.

At the surface $\dpi{120} \fn_cm (r=R)$

Electric field is $\dpi{120} \fn_cm \mathbf{\left [ E=\frac{1}{4\pi \epsilon_0}.\frac{Q}{R^2} \right ]}$

$\dpi{120} \fn_cm {\color{Red} \textbf{Case I :-}}$ When point P lies inside the shell

Draw a Gaussian surface as shown in Fig.

The flux through a small area dA at point P on the Gaussian surface is

$\dpi{120} \fn_cm d\Phi=\vec{E}.d\vec{A}=E dA\cos0^0=EdA$

∴ Flux through the total Gaussian surface is

$\dpi{120} \fn_cm \Phi=\int E.dA=E\int dA$

$\dpi{120} \fn_cm \Phi=E 4\pi r^2-----(1)$

From Gauss’s theorem, total flux through the Gaussian surface is

$\dpi{120} \fn_cm \Phi=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm \Phi=\frac{0}{\epsilon_0}$

$\dpi{120} \fn_cm \Phi=0-----(2)$

From (1) and (2)

$\dpi{120} \fn_cm E4\pi r^2=0$

$\dpi{120} \fn_cm \therefore \mathbf{\left [ E=\0 \right ]}$

i.e. inside the charged shell electric field is zero at any point.

Variation of the electric field with distance from the centre of the shell

Electric field due to uniformly charged insulating sphere

Consider a uniformly charged insulating sphere of charge Q and radius R.. We have to calculate the electric field at point P which is distance r from the centre of the sphere.

From symmetry, the magnitude of the electric field is the same at all points equidistance from the centre of the sphere

$\dpi{120} \fn_cm {\color{Red} \textbf{Case I :-}}$ When point P lies outside the shell

Draw a Gaussian surface as shown in the figure.

The flux through a small area dA at point P on the Gaussian surface is

$\dpi{120} \fn_cm d\Phi=\vec{E}.d\vec{A}=E dA\cos0^0=EdA$

∴ Flux through the total Gaussian surface is

$\dpi{120} \fn_cm \Phi=\int E.dA=E\int dA$

$\dpi{120} \fn_cm \Phi=E 4\pi r^2-----(1)$

From Gauss’s theorem, total flux through the Gaussian surface is

$\dpi{120} \fn_cm \Phi=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm \Phi=\frac{Q}{\epsilon_0}-----(2)$

From (1) and (2)

$\dpi{120} \fn_cm E.4\pi r^2=\frac{Q}{\epsilon_0}$

$\dpi{120} \fn_cm \therefore \mathbf{\left [ E=\frac{1}{4\pi \epsilon_0}.\frac{Q}{r^2} \right ]}$

i.e. Formula is such that, the total charge is concentrated at the centre of sphere.

Electric field due to uniformly charged infinite plane sheet

Consider an infinitely thin plane sheet of positive charge with a uniform charge density $σ$ . We have to calculate the electric field intensity at point P just in front of the sheet.

Draw a Gaussian surface as shown in the figure. This is a closed cylinder with flat area A.

From Fig, flux through a curved surface is zero because at any point electric field is perpendicular to the area vector. so flux is only the two flat surfaces of the Gaussian surface.

∴ Total flux through the Gaussian surface is

$\fn_cm \Phi=2EA\cos0^0=2EA-----(1)$

From Gauss’s theorem, total flux through the Gaussian surface is

$\fn_cm \Phi=\frac{Q_{inside}}{\epsilon_0}$

$\fn_cm \Phi=\frac{\sigma A}{\epsilon_0}-----(2)$

From (1) and (2)

$\fn_cm 2EA=\frac{\sigma A}{\epsilon_0}$

$\fn_cm \mathbf{\left [ E=\frac{\sigma}{2\epsilon_0} \right ]}$

If the total area of sheet is A and the total charge distributed is Q, then we can write electric field at point P is

$\fn_cm \mathbf{\left [ E=\frac{Q}{2A\epsilon_0} \right ]}$

Electric field due to charged metal plate

At point P,

$\fn_cm E=\frac{\sigma}{2\epsilon_0}+ \frac{\sigma}{2\epsilon_0}$

$\fn_cm \mathbf{\left [ E=\frac{\sigma}{\epsilon_0} \right ]}$

This formula applies to any shape and size of the conductor

Electric field due to hollow long cylinder with uniform charge distribution

Case 1:- Outside the cylinder

$\dpi{120} \fn_cm \oint \vec{E}.d\vec{A}=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm E2\pi rl=\frac{\lambda l}{\epsilon_0}$

$\dpi{120} \fn_cm \mathbf{\left [ E=\frac{\lambda}{2\pi \epsilon_0 r} \right ]}$

Case 2:- Inside the cylinder

$\dpi{120} \fn_cm \oint \vec{E}.d\vec{A}=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm E2\pi rl=\frac{0}{\epsilon_0}$

$\dpi{120} \fn_cm \mathbf{\left [ E=0 \right ]}$

Electric field due to a long non-conducting cylinder with uniform charge distributed through the volume.

Case 1:- Outside the cylinder

$\dpi{120} \fn_cm \oint \vec{E}.d\vec{A}=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm E2\pi rl=\frac{\pi R^2l}{\epsilon_0}$

$\dpi{120} \fn_cm \mathbf{\left [ E=\frac{\rho R^2}{2\epsilon_0 r} \right ]}$

Case 2:- Inside the cylinder

$\dpi{120} \fn_cm \oint \vec{E}.d\vec{A}=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm E2\pi rl=\frac{0}{\epsilon_0}$

$\dpi{120} \fn_cm \mathbf{\left [ E=\frac{\rho r}{2\epsilon_0 } \right ]}$

Case 3:- Surface of the cylinder (r=R)

$\dpi{120} \fn_cm \mathbf{\left [ E=\frac{\rho R}{2\epsilon_0 } \right ]}$

Electric field inside the cylindrical cavity

$\dpi{120} \fn_cm \vec{E}=\frac{\rho \vec{r}_1}{2\epsilon_0}+\frac{(-\rho) \vec{r}_2}{2\epsilon_0}$

$\dpi{120} \fn_cm =\frac{\rho (\vec{r}_1-\vec{r}_2)}{2\varepsilon_0}=\frac{\rho \vec{d}}{2\varepsilon_0}(constant))$

Electric field due to uniformly volume charged plane

Case 1:- Inside the plane

$\dpi{120} \fn_cm \oint \vec{E}.d\vec{A}=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm EA+EA+0=\frac{A2r\rho}{\varepsilon_0}$

$\dpi{120} \fn_cm 2EA=\frac{A2r\rho}{\varepsilon_0}$

$\dpi{120} \fn_cm \mathbf{\left [ E=\frac{\rho r}{\varepsilon_0} \right ]}$

Case 2:- outside the plane

$\dpi{120} \fn_cm \oint \vec{E}.d\vec{A}=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm 2EA=\frac{Ad\rho}{\varepsilon_0}$

$\dpi{120} \fn_cm \mathbf{\left [ E=\frac{\rho d}{2\varepsilon_0} \right ]}\;\;(constant)$

Exercise

(1) A thin copper wire 3 m long carries a net charge of 5.8 μC distributed uniformly. What is the electric field 8 cm from the wire? $\dpi{120} \fn_cm (4.8\times 10^5 N/C)$

(2) An infinite line charge produces a field of $\dpi{120} \fn_cm 9\times10^4 N/C$ at a distance of 2 cm. Calculate the linear charge density. $\dpi{120} \fn_cm (10^{-7}C/m)$

(3) An electron is revolving around a long line charge having charge density $\dpi{120} \fn_cm 2\times 10^{-8} C/m$ . Find the K.E of the electron. $\dpi{120} \fn_cm (2.88 \times 10^{-17}J)$

(4) Calculate the net force acting on the dipole. (0.7N left)

(5) Find the magnitude of the force exerted on the unit length of one due to the charge on the other. $\dpi{120} \fn_cm \left ( \frac{\lambda_1\lambda_2}{2\pi \epsilon_0 d} \right )$

(6) Find the electric field at A,B and C on both case

(7) A spherical shell of radius 5 cm is charged uniformly to 10μC. If the sphere is placed in air, find the electric field intensity at a point (a) 10 cm from the centre and (b) 4 cm from the centre $\dpi{120} \fn_cm (9\times 10^6 N/C, 0)$

(8) The nucleus of an atom of gold has a radius $\dpi{120} \fn_cm R=6.2 \times 10^{-15}m$ . Find the magnitude of the electric field (a) At the surface of the nucleus (b) At a distance of 2R from the centre of the nucleus. $\dpi{120} \fn_cm (3\times10^{21}N/C, 0.75\times10^{21} N/C)$

(9) A flat square sheet of charge of side 50 cm carries a uniform surface charge density**************. An electron 1.5 cm from a point near the centre of the sheet experiences a force of $\dpi{120} \fn_cm 1.8 \times10^{-12}N$ directed away from the sheet. Determine the total charge on the sheet. (-50 μC)

(10) A large plane sheet of charge having a surface charge density $\dpi{120} \fn_cm 5.0\times 10^{-6} C/m^2$ lies in the x-y plane. Find the electric flux through a circular area of radius 0.1 m, if the normal to the circular area makes an angle of 60° with the z-axis. $\dpi{120} \fn_cm (4.44\times10^3 Nm^2/C)$

(11) A particle of mass $\dpi{120} \fn_cm 9\times 10^{-5}\;gram$ is kept over a large horizontal sheet of charge density $\dpi{120} \fn_cm 5\times 10^{-5}C/m^2$ . What charge should be given to the particle, so that if released, it doesn’t fall? $\dpi{120} \fn_cm (3.12\times10^{-13}C)$

(12) Find the magnitude of electric field at a point x, from the common centre for (a) 0<x<a (b) a≤x<b (c)b≤x<∞ $\dpi{120} \fn_cm (0,\frac{kq}{x^2},\frac{k(q+Q)}{x^2})$

(13) Two conducting plates A and B are placed parallel to each other. A is given a charge q1 and B a charge q2. Prove that the charges on the inner surfaces are equal in magnitude and opposite sign.

(14) Find the charge appearing on (1), (2), (3) and (4)

Note:- Charges appearing on surfaces (2) and (3) are called bounded charges and charges appearing on (1) and (2) are called free charges.

(15) Find the charge appearing on each face. ( 2Q, -Q, Q, 3Q, -3Q, 2Q)

(16) Find the electric field at A, B, C and D. $\dpi{120} \fn_cm \left ( \frac{Q}{A\epsilon_0}left,0,\frac{2Q}{A\epsilon_0}left,\frac{Q}{A\epsilon_0}right \right )$

(17) Calculate the charges on each face.

(18) Find the charge on the outer surface of the rightmost plate. (-Q/2)

The behaviour of metallic conductors in an electric field

(1) The net electric field inside the conductor is zero

(2) The net charge inside the conductor is zero

From Gauss’s theorem

$\dpi{120} \fn_cm \oint \vec{E}.d\vec{A}=\frac{Q_{inside}}{\epsilon_0}$

$\dpi{120} \fn_cm \because$ Inside the conductor $\dpi{120} \fn_cm E=0$

$\dpi{120} \fn_cm \therefore \mathbf{\left [ Q_{inside}=0 \right ]}$

i.e., All extra charges reside on the outer surface of the conductor.

(3) At the surface of a charged conductor, the electric field must be normal to the surface at every point.

If the component of the electric field along the tangent to the surface were not zero, then the tangential component would exert forces on the surface charges causing them to move. But this is not possible because the conductor is in electrostatic equilibrium

(4) The magnitude of the electric field just outside a charged conductor is $\dpi{120} \fn_cm \sigma/\varepsilon_0$

(5) The electric potential is the same (constant) at the surface and inside a charged conductor.

(6) Electrostatic Shielding