Question

How much work does the electric field do in moving a -7.7μC charge from ground to a point whose potential is +65 V higher?

Answer 1

The potential energy of the charge in the field is \(U = q V\)

So the **change** in PE is \(\Delta U = q \Delta V = (-7.7 \times 10^{-6} C)(65V)\)

The work actually done by the field is \(W = - \Delta U\)

Answer 2

Or the better way IMHO, but depending upon what you know, is to take the given:

\(E = - \dfrac{dV}{dx}\)

And then to note that the work done by any force is:

\(W = \int F(x) ~ dx\)

Which in the case of an E field is:
\(= q\int E(x) ~ dx\)
\(= - q\int \dfrac{dV}{dx} ~ dx\)
\(\implies W = - q \Delta V\)