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Numerical

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.

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#### Solution

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.

Concept: Charging by Induction

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