Q.
Two point charges +8μC and +12μC repel each other with a force of 48N. When an additional charge of −10μC is given to each of these charges (the distance between the charges is unaltered) then the new force is
Given, q1=+8μC,q2=+12μC
and F1=48N
When an amount of charge −10μC is added to
both of the charges then, these becomes q1′=−2μC and q2′=+2μC
The force between two charges, F∝∣q1q2∣ ⇒F1∝∣q1q2∣ and F2∝∣∣q11q21∣∣
and F2F1=q1′q2′q1q2⇒F248=2×28×12 ⇒F2=2N