a. Two insulated charged copper spheres A and B have their centres separated by a distance o 0.5 m. What is the mutual force of electrostatic repulsion if the charge on each is `6.5xx10^(-7)C`? Assume that the radii of A and B are negligible compared to the distance fof separation.
b. What is the fore of repulsion if each sphere is charged double the above amount and the distance between them is halved?

Explanation

(a) Charge on sphere , r=50cm=0.5m
Force of repulsion between the two spheres,
`F=(q_(A)q_(B))/(4piepsilon_(0)r^(2))`
where , `epsilon_(0)=` Free space permittivity
`(1)/(4piepsilon_(0))=9xx10^(9)Nm^(2)C^(-2)`
`F=(9xx10^(9)xx(6.5xx10^(-7))^(2))/((0.5)^(2))`
`=1.52xx10^(-2)N`
Therefore , the force between the two sphere is `1.52xx10^(-2)N`.
(b) After doubling the charge , charge on sphere A,`q_(A)=` Charge on sphere `B,q_(B)=2xx6.5xx10^(-7)C=1.3xx10^(-6)C`
The distance between the sphere is halved.
`thereforer=(0.5)/(2)=0.25m`
Force of repulsion between the two spheres,
`F=(q_(A)q_(B))/(4piepsilon_(0)r^(2))`
`=(9xx10^(9)xx1.3xx10^(-6)xx1.3xx10^(-6))/((0.25)^(2))`
`=16xx1.52xx10^(-2)`
`=0.243N`
Therefore , the force between the two spheres is `0.243N`.