Question

Two similarly and equally charged identical metal spheres A and B repel each other with a force of $2\times {10}^{-5}N$. A third identical uncharged sphere C is touched with A and then placed at the midpoint between A and B. Find the net electric force on C :

• A. $4\times {10}^{-5}N$ away from sphere B
• B. $4\times {10}^{-5}N$ toward sphere B
• C. $2\times {10}^{-5}N$ away from sphere B
• D. $2\times {10}^{-5}N$ toward sphere B

Answer 1

The correct option is C $2\times {10}^{-5}N$ away from sphere B

Let the charge on each sphere be q then the force on each sphere is $F=\frac{kqq}{r^2}=\frac{k{q}^2}{r^2}=2\times {10}^{-5}.....\left(1\right)$

When sphere C is touched by sphere A, charge q gets equally divided between the two.

Thus, $F_{CA}=\frac{k\left(q/2\right)\left(q/2\right)}{r^2/4}=\frac{k{q}^2}{r^2}$

and $F_{CB}=\frac{k\left(q/2\right)\left(q\right)}{r^2/4}=\frac{2k{q}^2}{r^2}$

Net force $=F_{CB}-F_{CA}=\frac{2k{q}^2}{r^2}-\frac{k{q}^2}{r^2}=\frac{k{q}^2}{r^2}=2\times {10}^{-5}N$ (using (1))

As $F_{CB}>F_{CA}$ so net force is away sphere B.