Question

It takes 15 joules of work to bring 3.0 coulombs of positive charge from infinity to a point. What is the electric potential at this point in an electric field?

Answer 1

The work done to bring the charge from infinity to point A is equal to the variation of potential energy of the charge:
$W=U(A)-U(\infty )= q V(A)- q V(\infty)$
Using $V(\infty)=0$ as reference, we have:
$W= q V(A)$
from which we find the potential at point A:
$V(A) = \frac{W}{q}= \frac{15 J}{3.0 C}=5 V$