Question

If force f between two equal and opposite charge which are placed at a certain distance apart. If 25 %of one charge is transfered to the other then what is the force between them ?

Answer 1

force is unaltered.. because the magnitude of product of charges will remain same even after 25% is transferred from one charge to another charge

Answer 2

Answer: answer is 9f/16

Explanation: the key point here is that the two given charges are opposite and the force between them is given as

f=kq^2/r^2.......(1)

Let us consider q1 as+ve and q2 as-ve

Now the 25% charge transferred from q1 can be written as +q-q/4=3q/4

The 25% charge gained by q2 can be written as -q+q/4=-3q/4.

Now F'=|k(3q/4)(-3q/4)|/r^2

I have taken mod on rhs (not l)

From F'=9kq^2/16/r^2 .......(2)

From (1) equation (2) becomes

F'=9f/16