Question

The electrostatic force between two charges q1 and q2 separated by a distance r is F. If the charges are doubled and the separation between them is reduced to half, the electrostatic force will be

Answer 1

AnSweR :-

Initial force between the charges          F=r2kq1q2

Now distance between charges is doubled and both the charges are also doubled   i.e  q1′=2q1,        q2′=2q2         and        r′=2r

Thus new force    F′=(2r)2k(2q1)(2q2)=F

Answer 2

Answer:

Explanation:

F=9×10⁹×Q1×Q2/d²

If charges double and distance half then

F=9×10⁹×(2Q1)×(2Q2)/(d/2)²

= 9×10⁹×Q1×Q2×4/d²/4

But

F=9×10⁹×Q1×Q2/d²

So that

F= F×16