Question

The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become

• A. 7.20 N
• B. 11.25 N
• C. 22.50 N
• D. 45.00 N

Answer 1

The correct option is B 11.25 N

$F\propto \frac{1}{r^2}\Rightarrow \frac{F_1}{F_2}={\left(\frac{r_2}{r_1}\right)}^2\Rightarrow \frac{5}{F_2}={\left(\frac{0.04}{0.06}\right)}^2=F_2=11.25N$