Question

In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point

Answer 1

The difference in electric potential energy between the two points is
$\Delta U = q \Delta V$
where q is the magnitude of the charge and $\Delta V$ is the electric potential difference.

But for energy conservation, the difference in electric potential energy $\Delta U$ between the two points is equal to the work done to move the charge between A and B:
$W=\Delta U$
so we have
$W=q \Delta V$

and by substituting the numbers of the problem, we find the value of $\Delta V$:
$\Delta V = \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V$