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  4. 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 :-

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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 :-

Solution:

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, $F _{\text {finally }} \propto \frac{ Q }{2} \times \frac{ Q }{2}$
$F _{\text {initially }} \propto( Q )(2 Q )$
$\Rightarrow \frac{ F _{\text {finally }}}{ F _{\text {initially }}}=\frac{1}{8}$
$\Rightarrow F _{\text {finally }}=\frac{ F }{8}$