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  4. The magnitude of the two charges is doubled and the distance of their separation is also doubled. The electrostatic force between them will :

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Q. The magnitude of the two charges is doubled and the distance of their separation is also doubled. The electrostatic force between them will :

Electrostatic Potential and Capacitance

Solution:

$F_1 = \frac{1}{4 \, \pi \, \varepsilon_0} \frac{q_1q_2}{r^2}$ and $F_2 = \frac{1}{4 \, \pi \, \varepsilon_0} = \frac{(2q_1)(2q_2)}{(2r)^2} = F_1$