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. $\frac{3F}{4}$
• C. $\frac{F}{2}$
• D. $\frac{F}{4}$

Answer 1

The correct option is A F

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