Question

Two isolated conducting spheres ${S_1}$ and ${S_2}$ of radius $\dfrac{2}{3}R$ and $\dfrac{1}{3}R$ have $12\mu C$ and $ - 3\mu C$ charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on ${S_1}$ and ${S_2}$are respectively:
(A) $6\mu C$ and $3\mu C$
(B) $4.5\mu C$ on both
(C) $4.5\mu C$ and $ - 4.5\mu C$
(D) $3\mu C$ and $6\mu C$

Answer 1

Hint: In order to solve this question, we will use the concept that the sum of charges will remain the same before and after joining both spheres, and after joining the sphere through the wire, both spheres will have the same potential and hence by forming two equations we will solve for a charge on each sphere.

Formula used:
Potential on a sphere in terms of charge and radius is given by:
$V = \dfrac{{KQ}}{R}$
Where Q is charge, R is radius and K is constant.

Complete answer:
Let us draw the rough diagram of before and after joining the spheres as

So, the sum of charges will remain same after joining the sphere as
$
  {q_1} + {q_2} = 12 - 3 \\
  {q_1} + {q_2} = 9\mu C \to (i) \\
 $

Now, after joining the spheres, the potential remains the same, and using the formula $V = \dfrac{{KQ}}{R}$ for both sphere we get,
$
  \dfrac{{K{q_1}}}{{(\dfrac{2}{3}R)}} = \dfrac{{K{q_2}}}{{(\dfrac{1}{3})R}} \\
  {q_1} = 2{q_2} \to (ii) \\
 $

Now, solving equations (i) and (ii) we get,
$
  3{q_2} = 9 \\
  {q_2} = 3\mu C \\
 $
and
$
  {q_1} = 9 - 3 \\
   \Rightarrow {q_1} = 6\mu C \\
 $

So, the charges on spheres ${S_1}$ and ${S_2}$ are $6\mu C$ and $3\mu C$.

Hence, the correct answer is option (A) $6\mu C$ and $3\mu C$

Note: It should be remembered that whenever the two charge spheres are joined together by a wire they share a common potential but the charges are distributed among them according to their radius and thus depend upon the capacitance of the spheres. and $\mu C$ is the smaller unit of charge and it’s related as $1\mu C = {10^{ - 6}}C$.