Question

F is the force between two charges. If the distance between them is tripled, then the force between the charges will be :

• A. F
• B. $\frac{F}{3}$
• C. $\frac{F}{9}$
• D. $\frac{F}{27}$

Answer 1

Answer: (C)

$\frac{F}{9}$

$F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$ and

$F^{\prime }=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{(3r{)}^2}$

$=\frac{1}{9}(\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2})=\frac{F}{9}$