Question

How much work does the electric field do in moving a proton from a point with a potential of +145 v to a point where it is -55 v ? express your answer in joules?

Answer 1

The work done by the electric field in moving a charge is the negative of the potential energy difference between the two locations, which is the product between the magnitude of the charge q and the potential difference $\Delta V$:
$W=-\Delta U=-q\Delta V=-q (V_f -V_i)$
The proton charge is $q=1.6 \cdot 10^{-19}C$, and the two locations have potential of $V_i = +145 V$ and $V_f=-55 V$, therefore the work is
$W=-(1.6 \cdot 10^{-19}C)(-55 V-(+145 V))=3.2 \cdot 10^{-17}J$