Question

A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of 3Q, the new potential difference between the same two surfaces is nV. Find n.

Answer 1

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Answer 2

Let $R_1$ be the radius of the solid conducting sphere and $R_3$ be the outer radius of the hollow spherical shell. Let $R_2$ be the inner radius of the hollow spherical shell, where $R_1 < R_2 < R_3$.

Initially, the solid sphere has charge $Q$, and the hollow shell is uncharged (net charge is 0). Due to the charge $Q$ on the inner sphere, a charge $-Q$ is induced on the inner surface of the shell (at radius $R_2$). Since the total charge on the shell is 0, a charge $+Q$ must reside on the outer surface of the shell (at radius $R_3$).

Let $V(r)$ be the electric potential at a distance $r$ from the center. The potential at the surface of the solid sphere ($r=R_1$) is $V_1 = V(R_1)$. The potential at the outer surface of the hollow shell ($r=R_3$) is $V_3 = V(R_3)$. The potential difference is $V = V_1 - V_3$.

To find $V_1$, we sum the potentials at $R_1$ due to the charge $Q$ on the inner sphere, the induced charge $-Q$ on the inner surface of the shell, and the induced charge $+Q$ on the outer surface of the shell. Potential at $r=R_1$ due to charge $Q$ on sphere: $\frac{1}{4\pi\varepsilon_0} \frac{Q}{R_1}$ (potential at the surface of the sphere). Potential at $r=R_1$ due to charge $-Q$ at $R_2$: $\frac{1}{4\pi\varepsilon_0} \frac{-Q}{R_2}$ (potential inside a spherical shell is constant and equal to the potential on the surface). Potential at $r=R_1$ due to charge $+Q$ at $R_3$: $\frac{1}{4\pi\varepsilon_0} \frac{+Q}{R_3}$ (potential inside a spherical shell is constant and equal to the potential on the surface). So, $V_1 = \frac{1}{4\pi\varepsilon_0} \left(\frac{Q}{R_1} - \frac{Q}{R_2} + \frac{Q}{R_3}\right)$.

To find $V_3$, the potential at $r=R_3$: For $r \ge R_3$, the system of charges behaves like a single point charge located at the center with a total charge equal to the sum of all charges inside or on the shell, which is $Q + 0 = Q$. So, $V_3 = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R_3}$.

The initial potential difference is $V = V_1 - V_3$: $V = \frac{1}{4\pi\varepsilon_0} \left(\frac{Q}{R_1} - \frac{Q}{R_2} + \frac{Q}{R_3}\right) - \frac{1}{4\pi\varepsilon_0} \frac{Q}{R_3}$ $V = \frac{Q}{4\pi\varepsilon_0} \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.

Now, the shell is given a charge of $3Q$. The charge on the solid sphere remains $Q$. The net charge on the hollow spherical shell is now $3Q$. Due to the charge $Q$ on the inner sphere, a charge $-Q$ is still induced on the inner surface of the shell (at radius $R_2$). Since the total charge on the shell is $3Q$, the charge on the outer surface of the shell (at radius $R_3$) must be $3Q - (-Q) = 4Q$.

Let the new potentials be $V'_1$ at $r=R_1$ and $V'_3$ at $r=R_3$. The new potential difference is $V' = V'_1 - V'_3$.

To find $V'_1$, we sum the potentials at $R_1$ due to the charge $Q$ on the inner sphere, the induced charge $-Q$ on the inner surface of the shell, and the charge $4Q$ on the outer surface of the shell. $V'_1 = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R_1} + \frac{1}{4\pi\varepsilon_0} \frac{-Q}{R_2} + \frac{1}{4\pi\varepsilon_0} \frac{4Q}{R_3}$ $V'_1 = \frac{1}{4\pi\varepsilon_0} \left(\frac{Q}{R_1} - \frac{Q}{R_2} + \frac{4Q}{R_3}\right)$.

To find $V'_3$, the potential at $r=R_3$: For $r \ge R_3$, the system of charges behaves like a single point charge located at the center with a total charge equal to the sum of all charges, which is $Q + 3Q = 4Q$. So, $V'_3 = \frac{1}{4\pi\varepsilon_0} \frac{4Q}{R_3}$.

The new potential difference is $V' = V'_1 - V'_3$: $V' = \frac{1}{4\pi\varepsilon_0} \left(\frac{Q}{R_1} - \frac{Q}{R_2} + \frac{4Q}{R_3}\right) - \frac{1}{4\pi\varepsilon_0} \frac{4Q}{R_3}$ $V' = \frac{Q}{4\pi\varepsilon_0} \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.

Comparing $V'$ with $V$: $V = \frac{Q}{4\pi\varepsilon_0} \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $V' = \frac{Q}{4\pi\varepsilon_0} \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ So, $V' = V$.

The problem states that the new potential difference is $nV$. $nV = V'$ $nV = V$ Assuming $V \neq 0$ (which is true if $R_1 \neq R_2$), we can divide by $V$: $n = 1$.

The potential difference between the surface of the inner sphere and the inner surface of the shell depends only on the charge on the inner sphere. Adding charge to the outer shell changes the potential of both surfaces by the same amount, so their difference remains unchanged. Let the potential of the inner sphere be $V_{in}$ and the potential of the outer surface of the shell be $V_{out}$. $V = V_{in} - V_{out}$. In the initial case, $V_{in} = \frac{Q}{4\pi\varepsilon_0 R_1} - \frac{Q}{4\pi\varepsilon_0 R_2} + \frac{Q}{4\pi\varepsilon_0 R_3}$ and $V_{out} = \frac{Q}{4\pi\varepsilon_0 R_3}$. $V = \frac{Q}{4\pi\varepsilon_0} (\frac{1}{R_1} - \frac{1}{R_2})$. When the shell is given a charge $3Q$, the charge distribution is $Q$ on the inner sphere, $-Q$ on the inner surface of the shell, and $4Q$ on the outer surface of the shell. The new potential $V'_{in} = \frac{Q}{4\pi\varepsilon_0 R_1} - \frac{Q}{4\pi\varepsilon_0 R_2} + \frac{4Q}{4\pi\varepsilon_0 R_3}$. The new potential $V'_{out} = \frac{4Q}{4\pi\varepsilon_0 R_3}$. The new potential difference $V' = V'_{in} - V'_{out} = (\frac{Q}{4\pi\varepsilon_0 R_1} - \frac{Q}{4\pi\varepsilon_0 R_2} + \frac{4Q}{4\pi\varepsilon_0 R_3}) - \frac{4Q}{4\pi\varepsilon_0 R_3} = \frac{Q}{4\pi\varepsilon_0} (\frac{1}{R_1} - \frac{1}{R_2})$. So $V' = V$. Given $V' = nV$, we have $V = nV$, which implies $n=1$.