Question

A +3.0 nC charge Q is initially at rest a distance of 10 cm(r₁) from a +5.0 nC charge q fixed at the origin. Naturally, the Coulomb force accelerates Q away from q, eventually reaching 15 cm(r₂).

The charge Q is repelled by q, thus having work done on it and losing potential energy. What is the work done by the electric field between r₁ and r₂? What is the change in the potential energy of the two-charge system from r₁ to r₂? How much kinetic energy does Q have at r₂?

Answer 1

F = k Q q / r^2

and W = integral of F.dr from r1 to r2

W = (k Q q )(-1/r2 + 1/r1)

W = (9 x 10^9 x 5 x 10^-9 x 3 x 10^-9) (-1/0.15 + 1/0.10)

W = 4.5 x 10^-7 J .......Ans

change in PE = - Work done = - 4.5 x 10^-7 J ..........Ans

from energy conservation,

this loss of Pe will convert into KE of Q.

so Ke at r2 = 4.5 x 10^-7 J ......Ans