Question

Surface charge density of a sphare of a radius $10cm$ is $8.85\times {10}^{-8}c/m^2.$ Potential at the centre of the sphare is

• A. $442.5V$
• B. ${10}^{-3}$
• C. $1000V$
• D. $885V$

Answer 1

Answer: (C)

$\begin{array}{c}\sigma =8.85\times {10}^{-8}c/m^2\\ \Rightarrow \frac{Q}{4\pi R^2}=8.85\times {10}^{-8}c/m^2\\ \therefore V=\frac{KQ}{R}=\frac{1}{4\pi R^2}\frac{Q\times R}{R\times R}\\ =\frac{8.85\times {10}^{-8}\times R}{E_o}\\ =\frac{8.85\times {10}^{-8}\times 10}{8.85\times {10}^{-12}\times 100}\\ =1000V\end{array}$

$\therefore$ Option $A$ is correct.