Name: Electric Field of a Charged Plane Sheet
Uploaded: 2019-07-19
Duration: 6 min 58 s
Description: Another distribution of charge. This time on an infinite plane sheet. Let’s see once again how Gauss’s law comes to the rescue in electric field calculations.

Question

Using Gauss’ theorem derive an expression for the electric intensity due to a charged thin spherical shell at

(i) external point,

(ii) surface point,

(iii) internal point.
Draw the graph showing the variation of electric intensity obtained as above with distance from the centre.

Accepted Answer

Expression for electric field intensity due to charged thin spherical shell.

(i) When point is considered outside the shell (external point outside the shell)

Let spherical shell has centred D and radius R charge q is uniformly distributed over the whole surface as surface charge density $σ = \frac{q}{A} = \frac{q}{4 π R^{2}}$

From centre O at distance $r > R$ , P be a point where electric field intensity is to be determined.

For this purpose distance $O P = r$ , is taken as a radius and a spherical Gaussian surface is drawn.

In this condition $\vec{E}$ and $\vec{d A}$ , are along same direction.

$∴$ Electric flux $= ϕ = \vec{E} \cdot \vec{d A} = E d A \cos 0 ° = E d A$ , and electric flux through the whole Gaussian surface $= ϕ = \oint \vec{E} \cdot \vec{d A}$ $= \oint E d A = E \cdot \oint d A$ $= E \times 4 π r^{2}$ .....(i)

Here charge q is enclosed by the Gaussian surface, so according to Gauss's theorem electric flux $= ϕ = \frac{q}{ϵ_{0}} . . . . . . . . . (ii)$

From equation (i) and (ii), we get $E \cdot 4 π r^{2} = \frac{q}{ϵ_{0}}$

$\Rightarrow E = \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}}$

$= \frac{q}{4 π R^{2}} \times \frac{R^{2}}{ε_{0} r^{2}} = \frac{σ R^{2}}{ε_{0} r^{2}} \Rightarrow E \propto \frac{1}{r^{2}}$

(ii) When point is situated on the surface of shell in this condition

$r = R$

$∴$ $E = \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}} =$ $\frac{1}{4 π ϵ_{0}} \cdot \frac{q}{R^{2}}$

$\Rightarrow E = \frac{q}{4 π R^{2}} . \frac{1}{ϵ_{0}} = \frac{σ}{ϵ_{0}}$

(iii) When point is situated inside the shell :-

When point is considered inside the spherical shell and by facing $O P = r$ , as a radius a Gaussian surface is drawn then electric flux through Gaussian surface.

$= ϕ = \oint \vec{E} \cdot \vec{d A} =$ $\oint E d A c o s 0^{\circ} = E \oint d A = E . 4 π r^{2}$ .....(i)

Here no charge is enclosed by Gaussian surface because charges are the outer part of shell i.e., $q_{i n} = 0$

According to Gauss's theorem:

$ϕ = \oint \vec{E} \cdot d \vec{A}$ $= \frac{q}{ε_{0}} = \frac{0}{ϵ_{0}} = 0$ ......(ii)

From (i) and (ii), we get $E \cdot 4 π r^{2} = 0 \Rightarrow E = 0$

Graphical representation:

(i) Outside point, $r > R$ (ii) Point in surface, $r = R$ , $E = \frac{σ}{ϵ_{0}}$ (iii) Point inside the shell $E = 0, r < R$

Using Gauss’ theorem derive an expression for the electric intensity due to a charged thin spherical shell at

(i) external point,

(ii) surface point,

(iii) internal point.
Draw the graph showing the variation of electric intensity obtained as above with distance from the centre.

Important Questions on Gauss' Theorem

Learn and Prepare for any exam you want !

Using Gauss’ theorem derive an expression for the electric intensity due to a charged thin spherical shell at (i) external point, (ii) surface point, (iii) internal point. Draw the graph showing the variation of electric intensity obtained as above with distance from the centre.

Important Questions on Gauss' Theorem

Learn and Prepare for any exam you want !

Using Gauss’ theorem derive an expression for the electric intensity due to a charged thin spherical shell at

(i) external point,

(ii) surface point,

(iii) internal point.
Draw the graph showing the variation of electric intensity obtained as above with distance from the centre.