Question

Two equally charged, identical metal spheres $A$ and $B$ repel each other with a force $F$. The spheres are kept fixed with a distance $r$ between them. A third identical, but uncharged sphere $C$ is brought in contact with $A$ and then placed at the mid-point of the line joining $A$ and $B$. The magnitude of the net electric force on $C$ is

• A. F
• B. 3F/4
• C. F/2
• D. F/4

Answer 1

Answer: (A)

Initially,

$F=k\frac{Q^2}{r^2}$ ......(i)
Finally,

Force on $C$ due to $A,F_A=\frac{k(Q/2{)}^2}{(r/2{)}^2}=\frac{k{Q}^2}{r^2}$
Force on $C$ due to $B,F_B=\frac{KQ(Q/2)}{(r/2{)}^2}=\frac{2K{Q}^2}{r^2}$
$\therefore$ Net force on $C,F_{net}=F_B-F_A=\frac{k{Q}^2}{r^2}=F$