Question

Two point charges $Q_{1}$ and $Q_{2}$ exert a force $F$ on each other when kept certain distance apart. If the charge on each particle is halved and the distance between the two particles is doubled, then the new force between the two particles would be

• A. $\frac{F}{2}$
• B. $\frac{F}{4}$
• C. $\frac{F}{8}$
• D. $\frac{F}{16}$

Answer 1

Answer: (D)

Given
$F=\frac{K Q_{1} Q_{2}}{r^{2}} \,\,\,...(i)$
if, $Q_{1}'=\frac{Q_{1}}{2}$
$Q_{2}'=\frac{Q_{2}}{2}$
$\&\, r=2 r$
Then $=F_{1}'=\frac{F}{16}$