Question #215127
Derive an expression for Electric field due uniformly charged solid sphere at a) outside point b) inside point using Gauss's law
1
Expert's answer
2021-07-08T13:23:12-0400

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=Qenϵ0E.dA= \frac{Q_{en}}{\epsilon_0}

E.(4πr2)=Qenϵ0E.(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 Qen=QQ_{en}=QE=kQr2E=\frac{kQ}{r^2} So, Eout​Eout=Q4πϵ0r2E_{out}=\frac{Q}{4\pi\epsilon_0 r^2} ii) For a point inside the shell :By Gauss's law,

E.A=Qenϵ0E.A=\frac{Q_{en}}{\epsilon_0}

E.(4πr2)=Qenϵ0E.(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, Qen=0Q_{en}=0 So 

E.dA=0E.dA=0


Ein=0E_{in}=0


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