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?

Respuesta :

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 [tex]\Delta V[/tex]:
[tex]W=-\Delta U=-q\Delta V=-q (V_f -V_i)[/tex]
The proton charge is [tex]q=1.6 \cdot 10^{-19}C[/tex], and the two locations have potential of [tex]V_i = +145 V[/tex] and [tex]V_f=-55 V[/tex], therefore the work is
[tex]W=-(1.6 \cdot 10^{-19}C)(-55 V-(+145 V))=3.2 \cdot 10^{-17}J[/tex]