Question

The electric potential at the centre of hemisphere of radius $R$ having uniform surface charge density $\sigma$ is

• A. $\frac{\sigma R}{{ϵ}_0}$
• B. $\frac{2\sigma R}{{ϵ}_0}$
• C. $\frac{\sigma R}{2{ϵ}_0}$
• D. $\frac{\sigma R}{3{ϵ}_0}$

Answer 1

The correct option is C $\frac{\sigma R}{2{ϵ}_0}$

Consider a very small shaded element on the hemisphere which is subtended by an angle $d\theta$ with centre as shown in figure,

The area of shaded element is
$dA=2\pi rRd\theta$

From figure, $r=R\sin \theta$, so
$dA=2\pi \left(R\sin \theta \right)Rd\theta$
$\therefore dA=2\pi R^2\sin \theta d\theta$

Charge on this element,
$dq=\sigma \times 2\pi R^2\sin \theta d\theta$

Potential at the centre $\text{C}$ due to this element,
$dV=\frac{1}{4\pi {ϵ}_0}\frac{dq}{R}$

Substituting the value of $dq$, we get
$dV=\frac{1}{4\pi {ϵ}_0}\frac{\sigma 2\pi R^2\sin \theta d\theta }{R}$

$\Rightarrow dV=\frac{\sigma R}{2{ϵ}_0}\sin \theta d\theta$

Now the potential at $\text{C}$ due to whole hemisphere,
$\int dV=\frac{\sigma R}{2{ϵ}_0}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin \theta d\theta$

$\therefore V=\frac{\sigma R}{2{ϵ}_0}$

Hence, option (c) is the correct answer.