Question

A conducting sphere $\text{A}$ of radius $a$, with charge $Q$, is placed concentrically inside a conducting shell $\text{B}$ of radius $b$. $\text{B}$ is earthed and $\text{C}$ is the common center of $\text{A}$ and $\text{B}$. Study the following statements.

$\text{I.}$ The potential at a distance $r$ from $\text{C}$, where $a\le r\le b$, is $\frac{1}{4\pi {ϵ}_0}\left(\frac{Q}{r}\right)$
$\text{II.}$ The potential difference between $\text{A}$ and $\text{B}$ is $\frac{Q}{4\pi {ϵ}_0}(\frac{1}{a}-\frac{1}{b})$
$\text{III.}$The potential at a distance $r$ from $\text{C}$, where $a\le r\le b$, is $\frac{Q}{4\pi {ϵ}_0}(\frac{1}{r}-\frac{1}{b})$


Which of the following statements are correct?

• A. Only $\text{(I)}$ and $\text{(II)}$
• B. Only $\text{(II)}$ and $\text{(III)}$
• C. Only $\text{(I)}$ and $\text{(III)}$
• D. All

Answer 1

The correct option is B Only $\text{(II)}$ and $\text{(III)}$

Given,
shell $\text{B}$ is earthed so the potential on this shell will be zero.i.e, $V_B=0$

From the conservation of charge, $-Q$ will be induced on the inner surface of outer shell as shown in figure.


Now calculating the potential $V$ for $a\le r\le b$,

$V=\frac{kQ}{r}-(V_{inner}{)}_b....(1)$

Where, potential due to inner surface, $(V_{inner}{)}_b=\frac{k(-Q)}{b}$

Now, from $\left(1\right)$,

$V=\frac{kQ}{r}-\frac{kQ}{b}....\left(2\right)$

Substituting the value of $k=\frac{1}{4\pi {ϵ}_0}$, we get
$V=\frac{1}{4\pi {ϵ}_0}Q(\frac{1}{r}-\frac{1}{b})$

Potential at $\text{A}$,

$V_A=\frac{1}{4\pi {ϵ}_0}Q(\frac{1}{a}-\frac{1}{b})$

So, potential difference between $\text{A}$ and $\text{B}$ will be,

$\Rightarrow \Delta V_{AB}=V_A-V_B$

$\Rightarrow \Delta V_{AB}=\frac{Q}{4\pi {ϵ}_0}(\frac{1}{a}-\frac{1}{b})-0$

$\therefore \Delta V_{AB}=\frac{Q}{4\pi {ϵ}_0}(\frac{1}{a}-\frac{1}{b})$

Hence, statements $\left(\text{II}\right)$ and $\left(\text{III}\right)$ are correct and so we can say that option $\left(b\right)$ is the correct choice.