Answer:
The work done by the electric field in Figure 1 to move a positive charge q from A, the positive plate, higher potential, to B, the negative plate, lower potential, is
W = −ΔPE = −qΔV.
The potential difference between points A and B is
−ΔV = −(VB − VA) = VA − VB = VAB.
Entering this into the expression for work yields W = qVAB.
Work is W = Fd cos θ; here cos θ = 1, since the path is parallel to the field, and so W = Fd. Since F = qE, we see that W = qEd. Substituting this expression for work into the previous equation gives qEd = qVAB.
The charge cancels, and so the voltage between points A and B is seen to be
{
V
AB
=
E
d
E
=
V
AB
d
(uniform E − field only)
Step-by-step explanation:
The relationship between V and E for parallel conducting plates is E=Vd E = V d. From a physicist's point of view, either ΔV or E can be used to describe any charge distribution
