Question

The charge Q is uniformly distributed on the spherical shell of radius R then the Electric field on the surface of the shell

Answer 1

Given:

Charge Q has been uniformly distributed on a spherical shell of radius R .

To find:

Electrostatic Field Intensity on the surface of the shell.

Concept:

Since the question has clearly mentioned about a shell , we can say that it behaves like a conductor. And we can apply Gauss' Theorem to calculate the Electrostatic Field Intensity on the surface of the shell.

Calculation:

We shall consider a spherical Gaussian Surface around the thin shell.

Let Electrostatic Field Intensity vector be " E " and the area vector be " ds " , and $\theta$ be the angle between field vector and area vector

As per Gauss' Law :

$\therefore \: \displaystyle \: \int \vec E . \vec {ds} = \frac{Q}{ \epsilon_{0}}$

$= > \: \displaystyle \: \int E \times {ds} \times \cos( \theta) = \dfrac{Q}{ \epsilon_{0}}$

$= > \: \displaystyle \: \int E \times {ds} \times \cos(0 \degree) = \dfrac{Q}{ \epsilon_{0}}$

$= > \: \displaystyle \: E \times \int {ds} = \dfrac{Q}{ \epsilon_{0}}$

$= > \: E \times 4\pi {R}^{2} = \dfrac{Q}{ \epsilon_{0}}$

$= > \: E = \dfrac{Q}{4\pi {R }^{2} \epsilon_{0}}$

$= > \: E = \dfrac{Q}{4\pi \epsilon_{0} {R }^{2}}$

So final answer :

$\boxed{ \red{ \bold{ \huge{ E = \dfrac{Q}{4\pi \epsilon_{0} {R }^{2}} }}}}$