Question

A charge of 1.5 × 1015 coulombs moves from point A to a lower potential at point B. The electric field is uniform at 2.5 × 10-5 newtons/coulomb and the distance along the field lines between the two points is 1.7 × 10-2 meters. What is the work done?

Answer 1

The work done by the electric field is given by:
$W=q \Delta V$
where q is the charge, and $\Delta V$ is the potential difference between point A and B. But the potential difference is equal to
$\Delta V = E d$
where E is the intensity of the electric field, while d is the distance between the two points A and B.
Therefore, the work done by the field is
$W=q \Delta V =qEd=$
$=1.5\cdot10^{15}~C \cdot 2.5\cdot 10^{-5}~N/C \cdot 1.7\cdot 10^{-2}~m = 6.38\cdot 10^8~J$