Answer
a) 0
b) -11 J
c) -11 J
Work Step by Step
a)
Electric field and electric potential are rrealted as
Charge *Electric field = - $ \dfrac {\Delta V}{\Delta x}$
But in this case, the change in the potential is zero as the charge is moving in a direction perpendicular to the electric field.
b)
Change in potential = - qEd,
Where q represents the charge,
E represents the electric field, and
d represents the distance.
Change in potential = - (4.5 $\times 10^{-6} \times 4.5 \times 10^{5} \times 0.6$) = -11 J.
c)
The change in electric potential = change in potential for (0, 6.0m) + change in potential for (6.0m,0)
Change in potential = -11 J.
