3 ) A charge of +5nC is moved by a distance of 4 cm against a uniform electric field of magnitude 2 x 10^{12} NC^{-1} from A to B.

(a) Calculate the work done in moving the charge from A to B.

(b) The charge is now released and returns to A. Calculate the kinetic energy of the charge at A.

[ **here is the complete worksheet**: **electrostatics worksheet 1** ]

*Solution:*

Since the charge is positive we have to do work to move it from A to B (since this is against the field). This work is given by,

W = Fs= QEs

= (5 x 10^{-9})(2 x 10^{12})(0.04) J = 400 J

When released the charge moves under the influence of the electric field and returns to A.

Gain in kinetic energy of the charge at A = potential energy stored in the charge when it is moved to B = work done to move it from A to B = 400 J

So, Gain in kinetic energy of the charge at A = 400 J

[ **here is the complete worksheet**: **electrostatics worksheet 1** ]