electrostatics – Worksheet 1 problem solution (Q3)

3 ) A charge of +5nC is moved by a distance of 4 cm against a uniform electric field of magnitude 2 x 10¹² 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 10¹²)(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 ]