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

State Gauss's theorem in electrostatics. Use this theorem to derive expressions for electric field due to uniformly charged thin spherical shell at-a point (i) outside and (ii) inside the shell.

Accepted Answer

Gauss's theorem in electrostatics:

The total electric flux $(ϕ)$ through any closed surface $(S)$ in free space is equal to $\frac{1}{ε_{\circ}}$ times the total electric charge $(q)$ enclosed by the surface. That is, $ϕ = \oint_{S} \vec{E} . \vec{d s} = \frac{q}{ε_{\circ}}$ , here $ε_{\circ}$ is permittivity of free space.

Electric field due to uniformly charged thin spherical shell at a point outside the shell:

Here, $O$ is the centre of the spherical shell of radius $R$ , and charge enclosed by the spherical shell is $q$ with uniformed charge density $σ$ . Assuming the dotted circle of radius $r$ to be Gaussian surface.

Let a point $P$ is outside the spherical shell. The distance between the centre of the spherical shell and the point $P$ is $r$ .

Here, $r > R$ . Now according to Gauss' theorem:

$ϕ = \oint_{S} \vec{E} . \vec{d s} = \frac{q}{ε_{\circ}}$

$\Rightarrow ϕ = \oint_{S} E d s \cos θ = \frac{q}{ε_{\circ}}$ Here, $θ = 0 °$ as $\vec{E}$ and $\vec{d s}$ are along the same direction.

Thus, $ϕ = \oint_{S} E d s = \frac{q}{ε_{\circ}} \Rightarrow E \oint_{S} d s = \frac{q}{ε_{\circ}}$

Since, $\oint_{S} d s = 4 π r^{2}$ Substituting this in above equation.

$E (4 π r^{2}) = \frac{q}{ε_{\circ}} \Rightarrow E = \frac{1}{4 π ε_{\circ}} \frac{q}{r^{2}}$

Hence, the expression for electric field at a point outside the uniformly charged thin spherical shell is $E = \frac{1}{4 π ε_{\circ}} \frac{q}{r^{2}}$ .

Electric field due to uniformly charged thin spherical shell at a point inside the shell:

When point $P$ is inside the shell then $r < R$ and inside the spherical shell charge $q = 0 C$ .

Then, $E_{i n s i d e} = \frac{1}{4 π ε_{\circ}} \frac{q}{r^{2}} = \frac{1}{4 π ε_{\circ}} \frac{(0 C)}{r^{2}} = 0 N C^{- 1}$ .

State Gauss's theorem in electrostatics. Use this theorem to derive expressions for electric field due to uniformly charged thin spherical shell at-a point (i) outside and (ii) inside the shell.

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