Two identical metal balls with charges +2 Q and - Q are separated by some distance, and exert a force F on each other. They are joined by a conducting wire, which is then removed. The force between them will now be :-

Question

Two identical metal balls with charges +2 Q and - Q are separated by some distance, and exert a force F on each other. They are joined by a conducting wire, which is then removed. The force between them will now be :- (A) F (B) F/2 (C) F/4 (D) F/8

Tardigrade Admin · Accepted Answer

If two charged are joined by wire and then removed, then charge equally distributed on both.
So, finally,

q_{1} = \frac{Q}{2}

and

q_{2} = \frac{Q}{2}

So,

F \propto q_{1} q_{2}

So,

F_{finally} \propto \frac{Q}{2} \times \frac{Q}{2}

F_{initially} \propto (Q) (2 Q)

\Rightarrow \frac{F _{finally}}{F _{initially}} = \frac{1}{8}

\Rightarrow F_{finally} = \frac{F}{8}

Q. Two identical metal balls with charges +2Q and −Q are separated by some distance, and exert a force F on each other. They are joined by a conducting wire, which is then removed. The force between them will now be :-

F

F/2

F/4

F/8

If two charged are joined by wire and then removed, then charge equally distributed on both. So, finally, q1​=2Q​ and q2​=2Q​ So, F∝q1​q2​ So, Ffinally ​∝2Q​×2Q​ Finitially ​∝(Q)(2Q) ⇒Finitially ​Ffinally ​​=81​ ⇒Ffinally ​=8F​

Q. Two identical metal balls with charges $+ 2 Q$ and $- Q$ are separated by some distance, and exert a force $F$ on each other. They are joined by a conducting wire, which is then removed. The force between them will now be :-