Q. 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) $F/4$
Explanation
(a) Initially
[image:1_image378.emf]
$F = k\frac{Q^{2}}{r^{2}}$ …(i)
Finally
[image:1_image379.emf]
Force on $C$ due to $A,\ F_{A} = \frac{k(Q/2)^{2}}{(r/2)^{2}} = \frac{kQ^{2}}{r^{2}}$
Force on $C$ due to $B,\ F_{B} = \frac{KQ(Q/2)}{(r/2)^{2}} = \frac{2kQ^{2}}{r^{2}}$
$\therefore$ Net force on $C,\ F_{net} = F_{B} - F_{A} = \frac{kQ^{2}}{r^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.