show that E=V/d can you
explain
Answers
Answer:
i tried my best, hope it helps.
Explanation:
the definition of an electric field is F/q, where q is in the field, and the definition of electric potential is electrical potential energy divided by charge.
E = F/q
V = E/q
W = Fd
So by substitution,
W = Eqd
I want to get to E = V/d so I'll solve for E...
E = W/qd
So W/q is somehow equal to V?
So W/q = E/q And by the work energy theorem,
W = delta E, and the voltage in E = V/d is in fact a potential difference.
Explanation:
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
{VAB=EdE=VABd (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.
VOLTAGE BETWEEN POINTS A AND B
{
V
AB
=
E
d
E
=
V
AB
d
{VAB=EdE=VABd (uniform E − field only)
where d is the distance from A to B, or the distance between the plates.
