Question

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

• A. increases by 10%
• B. decreases by 20%
• C. decreases by 17%
• D. decreases by 31%

Answer 1

Answer: (D)

We know,
$F=\frac{k{q}_1{q}_2}{r^2}$ or $F\propto \frac{1}{r^2}$ (for same $q_1$ and $q_2$)
Now r is increased by $20^{\circ }$ and corresponding force is $F^{\prime }.$
$\therefore \frac{F}{F^{\prime }}=\frac{r^{\prime 2}}{r^2}={\left(\frac{\frac{120}{100}r}{r}\right)}^2={\left(\frac{12}{10}\right)}^2$
or $F^{\prime }={\left(\frac{10}{12}\right)}^{2}F$
So, percentage decrease in force
$=\frac{F-{\left(\frac{10}{12}\right)}^{2}F}{F}\times 100$
$=\frac{144-100}{144}=\frac{44}{144}\times 100\%=31\%$