Question

A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$ and $R$ $( R > r )$. If the surface charge densities on the two shells are equal, the electric potential at the common centre is :

• A. $\frac{1}{4 \pi \varepsilon_{0}} \frac{( R +2 r ) Q }{2\left( R ^{2}+ r ^{2}\right)}$
• B. $\frac{1}{4 \pi \varepsilon_{0}} \frac{( R + r )}{2\left( R ^{2}+ r ^{2}\right)} Q$
• C. $\frac{1}{4 \pi \varepsilon_{0}} \frac{( R + r )}{\left( R ^{2}+ r ^{2}\right)} Q$
• D. $\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 R+r)}{\left(R^{2}+r^{2}\right)} Q$

Answer 1

The Correct Option is C

 To find the electric potential at the common center of two concentric conducting thin spherical shells with equal surface charge densities, we can use the concept of electric potential due to a spherical shell.

1. Let the charge on the inner shell (radius \(r\)) be \(Q_1\) and the charge on the outer shell (radius \(R\)) be \(Q_2\).
2. The surface charge density for the inner shell is given by: \(\sigma_1 = \frac{Q_1}{4 \pi r^2}\)
3. The surface charge density for the outer shell is given by: \(\sigma_2 = \frac{Q_2}{4 \pi R^2}\)
4. Since the surface charge densities are equal, \(\sigma_1 = \sigma_2\): 

\[\frac{Q_1}{4 \pi r^2} = \frac{Q_2}{4 \pi R^2}\]

1. This simplifies to: 

\[Q_1 R^2 = Q_2 r^2\]

1. Let the total charge be \(Q\), hence \(Q_1 + Q_2 = Q\).
2. From the equation \(Q_1 R^2 = Q_2 r^2\) and \(Q_1 + Q_2 = Q\), solve simultaneously to find \(Q_1\) and \(Q_2\). By substituting \(Q_2 = Q - Q_1\) in \(Q_1 R^2 = Q_2 r^2\), solve to get:

\[Q_1 = \frac{Q \cdot R^2}{R^2 + r^2}\]\[Q_2 = \frac{Q \cdot r^2}{R^2 + r^2}\]

1. The electric potential at the center due to each shell is given by:
   • The potential at the center due to the inner shell is 

\[V_1 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q_1}{r}\]

• The potential at the center due to the outer shell is 

\[V_2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q_2}{R}\]

1. The total potential at the common center is the sum \(V = V_1 + V_2\): 

\[V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q_1}{r} + \frac{Q_2}{R} \right)\]

1. Substitute the values of \(Q_1\) and \(Q_2\): 

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

1. Simplifying, we get: 

\[V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{(R + r)Q}{R^2 + r^2}\]

Therefore, the electric potential at the common center is:

Correct Answer:

\(\frac{1}{4 \pi \varepsilon_0} \frac{(R + r)}{R^2 + r^2} Q\)