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

.(1)/(4 π 0) ((R+r) Q)/( (R^2+r^2 .) )

(1)/(4 π 0) ((R-r) Q)/( (R^2+r^2 ))

(1)/(4 π 0) ((R+r) Q)/(2 (R^2+r^2 ))

(1)/(4 π 0) ((R-r) Q)/(2 (R^2+r^2 ))

Electrostatics

JEE Main level, single correct, medium

221 of 311

Question 221 of 311

A charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R(R$ $\gt r$ ) such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

Show answer and solution

Answer: (C) $\left.\frac{1}{4 \pi \varepsilon_0} \frac{(R+r) Q}{\left(R^2+r^2\right.}\right)$

Let

q_1

and

q_2

be charge on two spheres of radius '

r

' and '

R

' respectively As,

q_1+q_2=\mathrm{Q}

and

\sigma_1=\sigma_2 \quad

[Surface charge density are equal]

\therefore \frac{q_1}{r \pi r^2}=\frac{q_2}{4 \pi R^2}

So,

q_1=\frac{Q r^2}{R^2+r^2}

and

q_2=\frac{Q R^2}{R^2+r^2}

Now, potential,

V=\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_1}{r}+\frac{q_2}{R}\right]

\begin{aligned} & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q r}{R^2+r^2}+\frac{Q R}{R^2+r^2}\right] \\ & =\frac{Q(R+r)}{R^2+r^2} \frac{1}{4 \pi \varepsilon_0} \end{aligned}

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A charge of total amount QQQ is distributed over two concentric hollow spheres of radii rrr and R(RR(RR(R >r\gt r>r ) such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

JEE Main question paper PDF

More Physics practice

A charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R(R$ $\gt r$ ) such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is