Question

How much work does the electric field do in moving a proton from a point with a potential of +155 v to a point where it is -65 v ? express your answer in joules?

Answer 1

The work done by the electric field is equal to the loss of electric potential energy of the proton in moving from its initial location to its final location:
$W=-\Delta U = -q \Delta V = -q (V_f -V_i)$
where $q=1.6 \cdot 10^{-19}C$ is the proton charge, $V_f = -65 V$ and $V_i=+155 V$ are the voltages in the final and initial locations. Substituting, we get
$W=-(1.6 \cdot 10^{-19}C)(-65 V-(+155 V))=3.5 \cdot 10^{-17}J$