Question

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

• A. Increases by $10$%
• B. Decrease by $20$%
• C. Decrease by $17$%
• D. Decrease by $31$%

Answer 1

Answer: (D)

Decrease by $31$%

Let the initial distance between the charges $q_1$ and $q_2$ be $r$.

Thus force acting between the charges initially $F=\frac{k{q}_1{q}_2}{r^2}$

New distance between the charges $r^{\prime }=1.2r$

Thus, new force acting between the charges $F^{\prime }=\frac{k{q}_1{q}_2}{(r^{\prime }{)}^2}$

$\therefore$ $F^{\prime }=\frac{k{q}_1{q}_2}{(1.2r{)}^2}=\frac{k{q}_1{q}_2}{1.44r^2}$

Percentage decrease in force $\frac{F-F^{\prime }}{F}\times 100=\frac{\frac{k{q}_1{q}_2}{r^2}-\frac{k{q}_1{q}_2}{1.44r^2}}{\frac{k{q}_1{q}_2}{r^2}}\times 100$

$\therefore$ $\frac{F-F^{\prime }}{F}\times 100=\frac{0.44}{1.44}\times 100=31$%