Question

A conducting sphere is enclosed by a hollow conducting shell. Initially, the inner sphere has a charge $Q$. While the outer one is uncharged. The potential difference between the two spherical surfaces is found to be $V$. Later on, the outer shell is given a charge $–4Q$. The new potential difference between the two surfaces.

• A. $1V$
• B. $V$
• C. $-2V$
• D. $2V$

Answer 1

The correct option is B

$V$

Step 1. Given data:

$a=$Radius of sphere $A$

$b=$Radius of sphere $B$

$V=$Potential difference between sphere $A\&B$

Step 2. Calculating potential difference of sphere $\mathit{A}\mathbf{}\mathbf{\&}\mathbf{}\mathit{B}$ for $\mathit{c}\mathit{a}\mathit{s}\mathit{e}\mathbf{}\mathbf{1}\mathbf{.}$

$V_A=\frac{KQ}{a}$

$V_B=\frac{KQ}{b}$

Potential difference between sphere $A\&B$,

$\begin{array}{rcl}{V}_A-V_B& =& \frac{KQ}{a}-\frac{KQ}{b}\\ & \Rightarrow & ∆V=KQ\left(\frac{1}{a}-\frac{1}{b}\right)\\ \therefore ∆V& =& V\end{array}$

Step 3. Calculating potential difference of sphere $\mathit{A}\mathbf{}\mathbf{\&}\mathbf{}\mathit{B}$ for $\mathit{c}\mathit{a}\mathit{s}\mathit{e}\mathbf{}\mathbf{2}$ when outer the sphere is charged by $\mathbf{–}\mathbf{}\mathbf{4}\mathbf{}\mathit{Q}$.

$V_A=\frac{KQ}{a}-\frac{4KQ}{b}$

$V_B=\frac{KQ}{b}-\frac{4KQ}{b}$

Now, the Potential difference between a sphere $A\&B$

$\begin{array}{rcl}{V}_A-V_B& =& \left(\frac{KQ}{a}-\frac{4KQ}{b}\right)-\left(\frac{KQ}{b}-\frac{4KQ}{b}\right)\\ & \Rightarrow & ∆V=\frac{KQ}{a}-\frac{4KQ}{b}-\frac{KQ}{b}+\frac{4KQ}{b}\\ & \Rightarrow & ∆V=\frac{KQ}{a}-\frac{KQ}{b}\end{array}$

$\begin{array}{rcl}& \Rightarrow & ∆V=KQ\left(\frac{1}{a}-\frac{1}{b}\right)\end{array}$

$\begin{array}{rcl}\therefore ∆V& =& V\end{array}$

The potential difference between case 1 and case 2 is the same, $\begin{array}{rcl}∆V& =& V\end{array}$

Hence, the correct option is (B).