What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Solution

Now, if each sphere is charged double, and the distance between them is halved then the force of repulsion according to Coulomb's law is:

F = k . \frac{2 q_{1} 2 q_{2}}{(r / 2)^{2}} F = 16 k . \frac{q_{1} q_{2}}{r^{2}} = 16 \times 1.5 \times 10^{- 2} = 24 \times 10^{- 2} F = 0.24 N .

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