Question

Two charged identical metal spheres $A$ and $B$ repel each other with a force of $3\times {10}^{-5}N$. Another identical uncharged sphere $C$ is touched with sphere $A$ and then it is placed midway between $A$ and $B$. Then the magnitude of net force on $C$ is :

• A. $1\times {10}^{-5}N$
• B. $3\times {10}^{-5}N$
• C. $2\times {10}^{-5}N$
• D. $5\times {10}^{-5}N$

Answer 1

The correct option is B $3\times {10}^{-5}N$

Let the charge on spheres A and B be $Q$ each initially and separation between them be $d$.

Force between them $F=\frac{k{Q}^2}{d^2}=3\times {10}^{-5}$ $N$

Now the sphere C is touched with sphere A, so both these spheres will share equal charge among them.

$\therefore$ New charge on the spheres A and C is $Q/2$ each.

Net force on C,

$F^{\prime }=\frac{kQ\left(Q/2\right)}{(d/2{)}^2}-\frac{k\left(Q/2\right)\left(Q/2\right)}{(d/2{)}^2}=\frac{k{Q}^2}{d^2}$

$⟹$ $F^{\prime }=3\times {10}^{-5}$ $N$