Question

Charge $Q$ coulombs is uniformly distributed throughout the volume of a solid hemisphere of radius $R$ meters. Then the potential at centre $O$ of the base of the hemisphere in volts is

• A. $\frac{1}{4\pi {\epsilon }_o}\frac{3Q}{2R}$
• B. $\frac{1}{4\pi {\epsilon }_o}\frac{3Q}{4R}$
• C. $\frac{1}{4\pi {\epsilon }_o}\frac{Q}{4R}$
• D. $\frac{1}{4\pi {\epsilon }_o}\frac{Q}{8R}$

Answer 1

The correct option is A $\frac{1}{4\pi {\epsilon }_o}\frac{3Q}{2R}$

$\begin{array}{c}potentialatadis\tan cer\left(r<R\right)isV\left(r\right)=\frac{Q}{8\pi {\in }_{0}R}\left(3R^2-r^2\right)\\ \therefore potentialatthecentreofthesphere\left(r=0\right)=\frac{3Q}{8\pi {\in }_{0}R}=\frac{1}{4\pi {\in }_0}\frac{3Q}{2R}\end{array}$

Hence,

option $\left(A\right)$ is correct answer.