Question

A solid of radius $R$ is charged to $Q$ uniformly. At what distance from the surface is the electrostatic potential half of the potential at the centre?

• A. $R$
• B. $\frac{R}{2}$
• C. $\frac{R}{3}$
• D. $2R$

Answer 1

Answer: (C)

$\frac{R}{3}$

Given a solid sphere of radius $R$ charged to $Q$ uniformly.

To find the distance from the surface at which electrostatic potential is half of that at the center, so we know potential outside the sphere $V_1=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{x}$ where $x$ is the distance from the center $(x>R)$ and,

Potential inside the sphere, $V_2=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{2R}[3-\frac{x^2}{R^2}]$

For center, put $x=0$

$V_o=\frac{1}{4\pi {\epsilon }_0}\frac{3Q}{2R}$

The condition given is $V_1=\frac{V_O}{2}$

So, $\frac{1}{4\pi {\epsilon }_0}\frac{Q}{x}=\frac{1}{2}\frac{1}{4\pi {\epsilon }_0}\frac{3Q}{2R}$

$\Rightarrow x=\frac{4R}{3}$

Distance from surface is $=x-R$

$=\frac{4R}{3}-R=\frac{R}{3}$