Question

Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10⁻⁷ C? The radii of A and B are negligible compared to the distance of separation.

Answer 1

Charge on sphere A, q_A = Charge on sphere B, q_B = 6.5 × 10⁻⁷ C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres,

`"F" = ("q"_"A""q"_"B")/(4piin_0r^2)`

Where,

∈₀ = Free space permittivity

`1/(4piin_0) = 9 xx 10^9  "N m"^2  "C"^-2`

∴ `"F" = (9 xx 10^9 xx (6.5 xx 10^-7)^2)/((0.5)^2)`

= 1.52 × 10⁻² N

Therefore, the force between the two spheres is 1.52 × 10⁻² N.