Question

Using Gauss’s law derive an expression for the electric field intensity due to a uniformly charged thin spherical shell at a point.
i) Outside the shell
ii) Inside the shell

Answer 1

Hint: We know that the gauss law of electrostatics, “the total flux linked with a closed surface is $\dfrac{1}{{{\varepsilon _0}}}$ times the charge enclosed by the closed surface”. Inside a spherical shell there is no charge i.e., Q = 0, putting this in gauss equation we can find the field intensity inside the shell and for outside the shell we will integrate the closed surface for spherical shell, putting this value in gauss equation we can find the field intensity outside the shell.

Complete step by step answer:
 According to Gauss law, we have
$\int {\overrightarrow E. d\overrightarrow S } = \dfrac{Q}{{{\varepsilon _0}}}$
The Electric Field due to a uniformly charged thin spherical shell at both points

Let us assume R as the radius of the shell and Q as the charge which is uniformly distributed on the surface.
 i) For a point outside the shell:
By Gauss's law,
​​$E \times 4\pi {r^2} = \dfrac{{{Q_{eq}}}}{{{\varepsilon _0}}}$
Here, r is the distance from centre of shell ( r > R) and charge enclosed by the surface ${Q_{eq}} = Q$
So, ${E_{out}} = E = \dfrac{Q}{{4\pi {r^2}{\varepsilon _0}}}$
 ii) For a point inside the shell:
 By Gauss's law,
$E \times 4\pi {r^2} = \dfrac{{{Q_{eq}}}}{{{\varepsilon _0}}}$
Here, r is the distance from centre of shell (r < R) and charge inside the shell is, ${Q_{eq}} = 0$
So, ${E_{in}} = E = 0$
Hence, the electric field intensity outside the shell = ${E_{out}} = E = \dfrac{Q}{{4\pi {r^2}{\varepsilon _0}}}$
And electric field intensity inside the shell = ${E_{in}} = E = 0$

Note:
Hence we know that the total charge within a volume can be calculated by the total electric field flux coming out of a closed surface surrounding the volume is known as Gauss’s law. It is useful for the calculation of the electrostatic field for a symmetric system.