Question

Each of the two point charges are doubled and their distance is halved. Force of interaction becomes n times, where n is

• A. 4
• B. 1
• C. 1/16
• D. 16

Answer 1

Answer: (D)

From the formula, $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$ $F=\frac{1}{4\pi {\epsilon }_0}\frac{2q_1\times 2q_2}{{(r/2)}^2}$ $=\frac{4}{1/4}F$ $\Rightarrow$ $F=16F$ $\Rightarrow$ $nF=16F$ $\therefore$ $n=16$