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  4. The electrostatic force between two charges of 10 μ C and 20 μ C at a distance of 2m is to be halved. This can be effected by

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Q. The electrostatic force between two charges of $10\,\mu C$ and $20\,\mu C$ at a distance of $2m$ is to be halved. This can be effected by

COMEDKCOMEDK 2004Electrostatic Potential and Capacitance

Solution:

Here, F = $k \frac{q_1q_2}{d^2}$ and F' = $k \frac{(2q_1)q_2}{(2d)^2} = \frac{1}{2}\frac{k q_1 q_2}{d^2} \, i.e. $ F' = $\frac{F}{2}$