Question

Potential difference between centre and surface of the non conducting sphere of radius $R$ and uniform volume charge density $\rho$ within it will be

• A. $\frac{\rho R^2}{6{ϵ}_0}$
• B. $\frac{\rho R^2}{4{ϵ}_0}$
• C. $\frac{\rho R^2}{3{ϵ}_0}$
• D. $\frac{\rho R^2}{2{ϵ}_0}$

Answer 1

The correct option is A $\frac{\rho R^2}{6{ϵ}_0}$

Given that,
Radius of sphere $=R$
Charge density $=\rho =\frac{\text{Charge}}{\text{Volume of sphere}}$
Let the total charge in sphere is $q$.
Then
$\rho =\frac{q}{\left(\frac{4}{3}\right)\pi R^3}$
$\therefore q=\frac{4}{3}\pi \rho R^3$

Electric potential at the surface of sphere is
$V_s=\frac{q}{4\pi {ϵ}_{0}R}$
Electric potential at the centre of sphere for a uniform solid sphere is
$V_c=\frac{3}{2}\times V_s$
$V_c=\frac{3q}{2\times 4\pi {ϵ}_{0}R}$
According to the question
$V_s-V_c=\frac{3}{2}(\frac{1}{4\pi {ϵ}_0}.\frac{q}{R})-\frac{1}{4\pi {ϵ}_0}.\frac{q}{R}$

$V_c-V_s=\frac{q}{8\pi {ϵ}_{0}R}$
Substituting the value of $q$, we have
$V_c-V_s=\frac{\rho R^2}{6{ϵ}_0}$

Hence, option (a) is correct.