Answer to Question #215127 in Electric Circuits for Shreesh NK

Question

Derive an expression for Electric field due uniformly charged solid sphere at a) outside point b) inside point using Gauss's law

Answer 1

Let, R be the radius of the shell and Q be the charge uniformly distributed on the surface. i) For a point  outside the shell :By Gauss's law,​$E.dA= \frac{Q_{en}}{\epsilon_0}$

$$E.(4\pi r^2)=\frac{Q_{en}}{\epsilon_0}$$

here r be the distance from centre of shell (r>R) and charge enclosed by surface $Q_{en}=Q$ ​$E=\frac{kQ}{r^2}$ So, Eout​$E_{out}=\frac{Q}{4\pi\epsilon_0 r^2}$ ii) For a point inside the shell :By Gauss's law,

$E.A=\frac{Q_{en}}{\epsilon_0}$

​$E.(4\pi r^2)=\frac{Q_{en}}{\epsilon_0}$ ​here r be the distance from centre of shell (r<R) and charge inside the shell, $Q_{en}=0$ So 

$E.dA=0$

$E_{in}=0$