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

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

Electrostatic Potential and Capacitance

Solution:

$F_m = \frac{1}{4 \, \pi \, \varepsilon_0 \, \varepsilon_r} \frac{q_1 q_2}{r^2} $ and $F_0 = \frac{1}{4 \, \pi \, \varepsilon_0} \frac{q_1q_2}{r^2}$
Therefore $F_m = \frac{F_0}{\varepsilon_r} = \frac{10\, N}{5}$ = 2 N