KCET 2015 Physics Solved Paper

Given, charge q1=+6µC=6×10−6C; charge q2=+9µC
=9×10−6C seperation = d
Force between charges is given as
When a charge of +3 μC is given to both and separation = d, then
The force between charges is given as
Dividing Eq. (1) by Eq. (2), we get
‌

F

F′

=‌

6×10−6×9×10−6

3×10−6×6×10−6

=3
⇒F′=

F

3

Therefore, the new force of repulsion is

F

3