The change in electric potential when the charge is moving through uniform electric field
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Answer:
E=−ΔVΔs E = − Δ V Δ s , where Δs is the distance over which the change in potential, ΔV, takes place. The minus sign tells us that E points in the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.
Answer:
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In the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and electric field. For example, a uniform electric field E is produced by placing a potential difference (or voltage) ΔV across two parallel metal plates, labeled A and B. (See Figure 1.)
Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a more fundamental relationship between electric potential and electric field. From a physicist’s point of view, either ΔV or E can be used to describe any charge distribution. ΔV is most closely tied to energy, whereas E is most closely related to force. ΔV is a scalar quantity and has no direction, while E is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by E below.) The relationship between ΔV and E is revealed by calculating the work done by the force in moving a charge from point A to point B.
But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an interesting special case.
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)
where d is the distance from A to B, or the distance between the plates in Figure 1. Note that the above equation implies the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus the following relation among units is valid: 1 N/C = 1 V/m.
The relationship between V and E for parallel conducting plates is
E
=
V
d
. (Note that ΔV = VAB in magnitude. For a charge that is moved from plate A at higher potential to plate B at lower potential, a minus sign needs to be included as follows: –ΔV = VA – VB = VAB.