Question

Two equal and opposite charges are placed at a certain
distance, the force between them is F. If 25% of one
charge is transferred to other, then the force between
them is -
(A) F
(B) 9F/16
(C) 15F/16
(D) 4F/5
​

Answer 1

F=kq²/r²

the changes p and q respectively

after the change ,

Q1=0.75p ; Q2 =1.25q

Net force=k(0.75p)(1.25q)/r²

=0.9375kq²r²

=0.9375F=15F/16

Answer 2

Answer:

9q/16

Explanation:

F=kq^2/r^2 ------(I)

now, as the 25% of charge is transfered

so, q1= q-25%ofq

q1= q-q/4

q1=3q/4

we know that the charges are equal and opposite charge

so, q2=-q + 25%of q

q2=-q+q/4

q2=-4q+q/ 4

q2=-3q/4

now, new force

F'= k।q1q2।/r^2

F'=k 3q/43q/4/r^2

F'=kq^2/16r^2

F'=F9/16 (using I)