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) = (W)/(q)= (15 J)/(3.0 C)=5 V`