Question

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

Answer: (B)

11.25 N

Answer 2

Initial separation between the charges = 0.06m

Final separation between the charges = 0.04m

Since $F \propto \frac{1}{r^{2}}$⇒ $\frac{F_{1}}{F_{2}} = \left( \frac{r_{2}}{r_{1}} \right)^{2}$⇒ $\frac{5}{F_{2}} = \left( \frac{0.04}{0.06} \right)^{2} = \frac{4}{9}$⇒ $F_{2} = 11.25N$