Question

Two concentric conducting thin shells are given charges $q_1$ and $q_2$ as shown in the figure. Potential difference between the shells is $V$. Then choose the correct option(s).

• A. If $q_1$ becomes twice potential difference between shells will become $2V$
• B. If both $q_1$ and $q_2$ becomes twice potential difference between shells will become $4V$
• C. If both the shells are connected by a conducting wire charge $q_1$ will flow through the wire
• D. If both the shells are connected by a conducting wire field outside the outer shell will not change

Answer 1

Answer: (A), (C), (D)

If $q_1$ becomes twice potential difference between shells will become $2V$, If both the shells are connected by a conducting wire charge $q_1$ will flow through the wire, If both the shells are connected by a conducting wire field outside the outer shell will not change

Answer 2

Let the radius of the inner shell be $r_1$ and the radius of the outer shell be $r_2$. The charge on the inner shell is $q_1$ and the charge on the outer shell is $q_2$.

The potential at the surface of the inner shell ($r=r_1$) is given by the sum of the potential due to the charge on the inner shell and the potential due to the charge on the outer shell.

Potential at $r=r_1$ due to $q_1$ is $V_{11} = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_1}$.

Potential at $r=r_1$ due to $q_2$ is $V_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2}$ (since $r_1 < r_2$, the potential inside the outer shell due to $q_2$ is constant and equal to the potential on the surface of the outer shell).

So, the potential of the inner shell is $V_1 = V_{11} + V_{21} = \frac{1}{4\pi\epsilon_0} \left(\frac{q_1}{r_1} + \frac{q_2}{r_2}\right)$.

The potential at the surface of the outer shell ($r=r_2$) is given by the sum of the potential due to the charge on the inner shell and the potential due to the charge on the outer shell.

Potential at $r=r_2$ due to $q_1$ is $V_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_2}$ (since $r_2 > r_1$, the potential outside the inner shell due to $q_1$ is the same as a point charge at the center).

Potential at $r=r_2$ due to $q_2$ is $V_{22} = \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2}$.

So, the potential of the outer shell is $V_2 = V_{12} + V_{22} = \frac{1}{4\pi\epsilon_0} \left(\frac{q_1}{r_2} + \frac{q_2}{r_2}\right)$.

The potential difference between the shells is $V = V_1 - V_2 = \frac{1}{4\pi\epsilon_0} \left(\frac{q_1}{r_1} + \frac{q_2}{r_2}\right) - \frac{1}{4\pi\epsilon_0} \left(\frac{q_1}{r_2} + \frac{q_2}{r_2}\right) = \frac{1}{4\pi\epsilon_0} \left(\frac{q_1}{r_1} - \frac{q_1}{r_2}\right) = \frac{q_1}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$.

Option 1: If $q_1$ becomes twice, $q_1' = 2q_1$, and $q_2$ remains unchanged. The new potential difference is $V' = \frac{q_1'}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right) = \frac{2q_1}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right) = 2V$. This option is correct.

Option 2: If both $q_1$ and $q_2$ become twice, $q_1' = 2q_1$ and $q_2' = 2q_2$. The new potential difference is $V' = \frac{q_1'}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right) = \frac{2q_1}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right) = 2V$. This option states that the potential difference becomes $4V$, which is incorrect.

Option 3: If both the shells are connected by a conducting wire. When connected by a conducting wire, the shells form a single conductor, and charge will redistribute such that the potential on both shells is equal. Since they are connected, they are at the same potential. In a conductor, charge resides on the outer surface. Thus, the charge from the inner shell will flow to the outer shell. The total charge on the system is $q_1 + q_2$. After connecting, the charge on the inner shell becomes 0, and the charge on the outer shell becomes $q_1 + q_2$. The charge that flowed through the wire from the inner shell to the outer shell is $q_1 - 0 = q_1$. This option is correct.

Option 4: If both the shells are connected by a conducting wire, field outside the outer shell will not change. Before connecting, the electric field at a distance $r > r_2$ from the center is due to the total charge $q_1 + q_2$ enclosed within a spherical surface of radius $r$. By Gauss's law, $E \cdot 4\pi r^2 = \frac{q_1 + q_2}{\epsilon_0}$, so $E = \frac{1}{4\pi\epsilon_0} \frac{q_1 + q_2}{r^2}$. After connecting, the total charge remains $q_1 + q_2$, and this charge resides on the outer shell. The electric field at a distance $r > r_2$ from the center is still due to the total charge $q_1 + q_2$ enclosed within a spherical surface of radius $r$. By Gauss's law, $E' \cdot 4\pi r^2 = \frac{q_1 + q_2}{\epsilon_0}$, so $E' = \frac{1}{4\pi\epsilon_0} \frac{q_1 + q_2}{r^2}$. Thus, the electric field outside the outer shell does not change. This option is correct.