Question

The electrostolice force between
200 microcoulomb and 500 micrcoulomb placed in free space is 5gF. Find the
distance between two changes.​

Answer 1

Answer:

charge = 500 μC = 500 * 10^-6 C = 5 * 10^-4 C, => The electrostatic force between these two charges: F = 5 gf = 5 * 10^-3 kgf = 5 * 10^-2 N. Therefore, the distance between the two charges is 1.34 * 10^2 m.