Question

Two charges $q_1$ and $q_2$, are separated by a distance r and apply a force F to each other. If both charges are doubled and the distance between them is also doubled, find out the new force between them ?

• A. $\frac{1}{2}F$
• B. $\frac{1}{3}F$
• C. $F$
• D. $8F$
• E. $16F$

Answer 1

The correct option is C $F$

Initial force between the charges $F=\frac{k{q}_1{q}_2}{r^2}$

Now distance between charges is doubled and both the charges are also doubled i.e $q_1^{\prime }=2q_1$, $q_2^{\prime }=2q_2$ and $r^{\prime }=2r$

Thus new force $F^{\prime }=\frac{k\left(2q_1\right)\left(2q_2\right)}{(2r{)}^2}=F$