Question

Force between two charges when placed in free space is 10N. If they are in a medium of relative permittivity 5, the force between them will be

• A. 2N
• B. 50N
• C. 0.5N
• D. None

Answer 1

Answer: (A)

2N

Answer 2

$F_{m} = \frac{1}{4\pi\varepsilon_{0}} \times \frac{1}{\varepsilon_{r}} \times \frac{q_{1}q_{2}}{r^{2}}\therefore F_{m} = \frac{F_{0}}{\varepsilon_{r}} = \frac{10N}{5} = 2N$