Question

Distance between two point charges is increased by 20%. Force of interaction between the charges

• A. increases by 10%
• B. decreases by 20%
• C. decreases by 17%
• D. decreases by 31%

Answer 1

The Correct Option is D

We know,
$F = \frac{kq_1 q_2}{r^2}$ or $F \propto \frac{1}{r^2}$ (for same $q_1$ and $q_2$)
Now r is increased by $20^{\circ}$ and corresponding force is $F'.$
$\therefore \, \frac{F}{F'} =\frac{ r'^{2}}{r^{2}} = \left(\frac{\frac{120}{100}r}{r}\right)^{2} = \left(\frac{12}{10}\right)^{2} $
or $ F' = \left(\frac{10}{12}\right)^{2} F $
So, percentage decrease in force
$= \frac{F -\left(\frac{10}{12}\right)^{2}F}{F} \times 100$
$ = \frac{144 - 100}{144} = \frac{44}{144} \times 100\% = 31\%$