Q 1.12(a):  Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5\times 10^{-7}C? The radii of A and B are negligible compared to the distance of separation.

Answers (1)

Since, the radii of the spheres A and B are negligible compared to the distance of separation, we consider them as point object.

Given,

Charge on each of the spheres = 6.5\times 10^{-7}C

and Distance between them, r = 50 cm = 0.5 m

We know,

F = k\frac{q_{1}q_{2}}{r^2}

Therefore, the mutual force of electrostatic repulsion(Since they have same sign of charge)

F = 9\times10^9Nm^{2}C^{-2}\times\frac{(6.5\times10^{-7}\ C)^2}{(0.5\ m )^2} = 1.5\times10^{-2} N

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