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 $r_1=3 \mathrm{~cm}$ and $r_2=2$ $\mathrm{cm}$. Before they are connected, $s_1$ carries charge of 10 units. The electric potential at the point which is at a distance $4 \mathrm{~cm}$ from the centre of $S_1$ and a distance $3 \mathrm{~cm}$ from the centre of $S_2$ is

• Correct answer
  A

  $\frac{1}{4 \pi \varepsilon_0} \frac{17}{6}$
• B

  $\frac{1}{4 \pi \varepsilon_0} \frac{3}{2}$
• C

  $\frac{1}{4 \pi \varepsilon_0} \frac{1}{6}$
• D

  $\frac{1}{4 \pi \varepsilon_0} \frac{17}{12}$

Solution

When spheres are connected, charges are redistributed such that potential on their surfaces are same.
So, we have $$ \begin{aligned} q_1+q_2 & =10 \\ \frac{k q_1}{r_1} & =\frac{k q_2}{r_2} \Rightarrow \frac{q_1}{3}=\frac{q_2}{2} \end{aligned} $$ 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 $4 \mathrm{~cm}$ from $s_1$ and $3 \mathrm{~cm}$ from $s_2$ is $$ \begin{aligned} V & =\frac{k q_1}{r_1}+\frac{k q_2}{r_2}=k \frac{6}{4}+k \frac{4}{3} \\ & =k\left(\frac{3}{2}+\frac{4}{3}\right)=k \frac{17}{6} V=\frac{1}{4 \pi \varepsilon_0} \frac{17}{6} V \end{aligned} $$

Hence, the correct answer is option: 'A'.