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. 44577
• D. 16

Answer 1

Answer: (D)

16

Answer 2

From the formula, $F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ $F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2{{q}_{1}}\times 2{{q}_{2}}}{{{(r/2)}^{2}}}$ $=\frac{4}{1/4}F$ $\Rightarrow$ $F=16F$ $\Rightarrow$ $nF=16F$ $\therefore$ $n=16$