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

Given `r=0.5m, q_(A)=q_(B)=6.5xx10^(-5)C`
a. We know that `F=(1/(4 pi epsilon_(0)))(q_(1)q_(2))/(r^(2))`
Force of respulsio `F=(9xx10^(9)xx(6.5xx10^(-5))^(-2))/((0.5)^(2))N`
i.e. `F=1.521xx10^(-2)N`
b. Givne `q_(A)^(')=q_(B)^(')=2xx6.5xx10^(-5)C=13.0xx10^(-5)C`
`r'=4/2=0.25m`
i.e.`F'=(1/( 4 pi epsilon_(0)))((2q_(A))(2q_(B)))/((r//2)^(2))`
i.e. `F'=16F=16xx1.521xx10^(-2)N=24.336xx10^(-2)N=0.24N`
The new force is sixteen times the old and is repulsive.