Question

(a) 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 `6.5xx10^(-7) C` ? The radius of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved ?

Answer 1

(a) 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))/(4pi epsi_(0)r^(2))`
Where,
`epsi_(0)` = Free space permittivity
`(1)/(4 pi epsi_(0)) = 9 xx 10 ^(9) Nm^(2) C^(-2)`
` therefore F = ( 9xx 10^(9) xx (6.5 xx 10 ^(-7))^(2))/((0.5)^(2)) `
`= 1.52 xx 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, `q_(A)` = Charge on sphere B, `q_(B) = 2 × 6.5 × 10^(−7) C = 1.3 × 10^(−6) C`
The distance between the spheres is halved.
`therefore r =( 0.5)/(2)= 0.25 m`
`= (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.