Answer
a) $\Delta U=-4.37mJ$
b) $\Delta U=4.37mJ$
c) $\Delta U=0$
Work Step by Step
(a) We can find the change in electric potential energy as
$\Delta U=-q_{\circ}Ed$
We plug in the known values to obtain:
$\Delta U=-(12.5\times 10^{-6})(6350)(0.0550)$
$\Delta U=-4.37mJ$
(b) We can find the change in electric potential energy as
$\Delta U=-q_{\circ}Ed$
We plug in the known values to obtain:
$\Delta U=-(12.5\times 10^{-6})(6350)(-0.0550)$
$\Delta U=4.37mJ$
(c) In the positive y direction, the charge moves perpendicular to the field and thus the change in the electric field is zero (that is, $\Delta U=0$).
