First slide

Electrostatic potential

Question

A charge Q is distributed over two concentric hollow spheres of radii r and R (>r) such that the surface densities are equal. The potential at the common centre is

Easy

A

$\frac{Q (R^{2} + r^{2})}{4 {πε}_{0} (R + r)}$
B

$\frac{Q}{R + r}$
C

zero
D

$\frac{Q (R + r)}{4 {πε}_{0} (R^{2} + r^{2})}$

Solution

Let the charges on spheres of radii r and R be q₁ and q₂ respectively. Then Q = q₁+ q₂. The potential at the centre will be
$V = \frac{1}{4 {πε}_{0}} [\frac{q_{1}}{r} + \frac{q_{2}}{R}]$
As surface densities are equal, hence
$σ = \frac{q_{1}}{4 {πr}^{2}} = \frac{q_{2}}{4 {πR}^{2}} or q_{2} = q_{1} (R^{2} / r^{2})$
$\begin{aligned} ∴ V = \frac{1}{4 {πε}_{0}} [\frac{q_{1}}{r} + q_{1} \frac{R}{r^{2}}] \\ = \frac{1}{4 {πε}_{0}} \times \frac{q_{1}}{r^{2}} [r + R] \end{aligned}$
Now, $q_{1} + q_{2} = Q or q_{1} + q_{1} \frac{R^{2}}{r^{2}} = Q$
$∴ \frac{q_{1}}{r^{2}} = \frac{Q}{(R^{2} + r^{2})}$
So, $V = \frac{Q}{4 {πε}_{0}} [\frac{R + r}{R^{2} + r^{2}}]$

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