Question

The force acting between two point charges kept at a certain distance is $5$N. Now the magnitudes of charges are doubled and distance between them is halved, the force acting between them is _________ N.

• A. $5$
• B. $20$
• C. $40$
• D. $80$

Answer 1

The correct option is D $80$

The force acting between two charges $F=\frac{k{q}_1{q}_2}{r^2}$

where $q_1$ and $q_2$ are charges and $r$ is distance between $q_1$ and $q_2$

Let:

$q_1=1C$

$q_2=1C$

$F=\frac{k{q}_1{q}_2}{r^2}=5N$

Now

$q_1^{new}=2q_1$

$q_2^{new}=2q_2$

$r^{new}=\frac{1}{2}r$

$F^{new}=\frac{k{q}_1^{new}{q}_2^{new}}{r_{new}^2}=\frac{k2q_1\times 2q_2}{\frac{r^2}{4}}$

$F^{new}=16\frac{k{q}_1\times q_2}{r^2}=16\times 5=80$

Hence the correct option is (D).