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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.
- What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Solution
(a) 
Charge on sphere A, qA = Charge on sphere B, qB = 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.
(b) After doubling the charge, charge on sphere A, qA = Charge on sphere B, qB = 2 × 6.5 × 10−7 C = 1.3 × 10−6 C
The distance between the spheres is halved.
∴ `"r" = 0.5/2`
= 0.25 m
Force of repulsion between the two spheres,
`F = ("q"_"A" "q"_"B")/(4 pi ∈_0 r^2)`
= `(9 xx 10^9 xx 1.3 xx 10^-6 xx 1.3 xx 10^-6)/(0.25)^2`
= 16 × 1.52 × 10-2
= 0.243 N
Therefore, the force between the two spheres is 0.243 N.
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