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:
[tex]W=-\Delta U = -q \Delta V = -q (V_f -V_i)[/tex]
where [tex]q=1.6 \cdot 10^{-19}C[/tex] is the proton charge, [tex]V_f = -65 V[/tex] and [tex]V_i=+155 V[/tex] are the voltages in the final and initial locations. Substituting, we get
[tex]W=-(1.6 \cdot 10^{-19}C)(-65 V-(+155 V))=3.5 \cdot 10^{-17}J[/tex]
