Question

A charge Q is distributed over the two concentric hollow spheres of radii \(r\) and \(R\) (\(R>r\)) such the surface densities are equal. The potential at the common centre is {\(K=\frac{Q}{4\pi \epsilon _{0}}\), \(\epsilon _{0}\) = permittivity of free space}

• A. \(\frac{KR}{r}\)
• B. \(\frac{K(r+R)}{(R^{2}+r^{2})}\)
• C. \(\frac{K(R^{2}+r^{2})}{(R+r)}\)
• D. \(\frac{K(r+R)}{(R^{3}+r^{3})}\)

Hint:

Equal surface densities mean the charges are in the ratio of the squares of the radii; inside a shell the potential equals its surface value.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Equal surface density \(\sigma\) means \(Q_1\propto r^2\) and \(Q_2\propto R^2\), with \(Q_1+Q_2=Q\).

Step 2: Charges
\(Q_1=\dfrac{Qr^2}{r^2+R^2}\) (inner) and \(Q_2=\dfrac{QR^2}{r^2+R^2}\) (outer).

Step 3: Potential at the centre
Inside a shell the potential is constant, so \(V=\dfrac{Q_1}{4\pi\epsilon_0 r}+\dfrac{Q_2}{4\pi\epsilon_0 R}\).

Step 4: Result
\(V=K\left(\dfrac{r}{r^2+R^2}+\dfrac{R}{r^2+R^2}\right)=\dfrac{K(r+R)}{R^2+r^2}\).

Final Answer:
The potential at the centre is K(r + R) divided by (R squared + r squared). \[ \boxed{\text{(B) }\dfrac{K(r+R)}{(R^{2}+r^{2})}} \]