question_answer

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                                [UPSEAT 2004; DCE 2005]

A)                    F                                        

B)                    3F/4           

C)                    F/2                                    

D)                    F/4

Correct Answer: A

Solution :

             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, \[{{F}_{B}}=\frac{KQ(Q/2)}{{{(r/2)}^{2}}}=\frac{2K{{Q}^{2}}}{{{r}^{2}}}\]            \ Net force on C, \[{{F}_{net}}={{F}_{B}}-{{F}_{A}}=\frac{k{{Q}^{2}}}{{{r}^{2}}}=F\]