Question

A conducting sphere $S_1$ of radius $r_1$ is connected by a conducting wire to another conducting sphere $S_2$ of radius $r_2$, where $r_1=3cm$ and $r_2=2cm$. Before they are connected, $S_1$ carries charge of $10$ units. The electric potential at the point which is at a distance $4cm$ from the centre of $S_1$ and a distance $3cm$ from the centre of $S_2$ is

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{17}{6}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{3}{2}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{1}{6}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{17}{12}$

Answer 1

Answer: (A)

When spheres are connected, charges are redistributed such that potential on their surfaces
are same.

So, we have
$q_1+q_2=10$
and $\frac{k{q}_1}{r_1}=\frac{k{q}_2}{r_2}$
$\Rightarrow \frac{q_1}{3}=\frac{q_2}{2}$
Hence, $q_1=6$ units and $q_2=4$ units.
Now, potential at some point $P$ distant $4cm$ from $s_1$ and $3cm$ from $s_2$ is
$V=\frac{k{q}_1}{r_1}+\frac{k{q}_2}{r_2}=k\frac{6}{4}+k\frac{4}{3}$
$=k\left(\frac{3}{2}+\frac{4}{3}\right)=k\frac{17}{6}V$
$=\frac{1}{4\pi {\epsilon }_0}\frac{17}{6}V$