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V P = R P E · d l .

From our previous discussion of the potential energy of a charge in an electric field, the result is independent of the path chosen, and hence we can pick the integral path that is most convenient.

Consider the special case of a positive point charge q at the origin. To calculate the potential caused by q at a distance r from the origin relative to a reference of 0 at infinity (recall that we did the same for potential energy), let P = r and R = , with d l = d r = r ^ d r and use E = k q r 2 r ^ . When we evaluate the integral

V P = R P E · d l

for this system, we have

V r = r k q r 2 r ^ · r ^ d r ,

which simplifies to

V r = r k q r 2 d r = k q r k q = k q r .

This result,

V r = k q r

is the standard form of the potential of a point charge. This will be explored further in the next section.

To examine another interesting special case, suppose a uniform electric field E is produced by placing a potential difference (or voltage) Δ V across two parallel metal plates, labeled A and B ( [link] ). Examining this situation 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.

The figure shows electric field between two plates (A and B) with opposite charges. The plates are separated by distance d and have a potential difference V subscript AB. A positive charge q is located between the plates and moves from A to B.
The relationship between V and E for parallel conducting plates is E = V / d . (Note that Δ V = V A B 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: Δ V = V A V B = V A B . )

From a physicist’s point of view, either Δ V or E can be used to describe any interaction between charges. However, Δ V is a scalar quantity and has no direction, whereas E is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field, a scalar quantity, is represented by E .) The relationship between Δ V and E is revealed by calculating the work done by the electric force in moving a charge from point A to point B . But, as noted earlier, arbitrary charge distributions require calculus. We therefore look at a uniform electric field as an interesting special case.

The work done by the electric field in [link] to move a positive charge q from A , the positive plate, higher potential, to B , the negative plate, lower potential, is

W = Δ U = q Δ V .

The potential difference between points A and B is

Δ V = ( V B V A ) = V A V B = V A B .

Entering this into the expression for work yields

W = q V A B .

Work is W = F · d = F d cos θ ; here cos θ = 1 , since the path is parallel to the field. Thus, W = F d . Since F = q E , we see that W = q E d .

Substituting this expression for work into the previous equation gives

q E d = q V A B .

The charge cancels, so we obtain for the voltage between points A and B

V A B = E d E = V A B d } ( uniform E -field only )

where d is the distance from A to B , or the distance between the plates in [link] . Note that this equation implies that 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 .

Furthermore, we may extend this to the integral form. Substituting [link] into our definition for the potential difference between points A and B , we obtain

Questions & Answers

calculate ideal gas pressure of 0.300mol,v=2L T=40°c
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neither vector nor scalar
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18.51micro Coulombs
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Under which topic
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Two equal positive charges are repelling each other. The force on the charge on the left is 3.0 Newtons. Using your notes on Coulomb's law, and the forces acting on each of the charges, what is the force on the charge on the right?
Nya Reply
Using the same two positive charges, the left positive charge is increased so that its charge is 4 times LARGER than the charge on the right. Using your notes on Coulomb's law and changes to the charge, once the charge is increased, what is the new force of repulsion between the two positive charges?
A mass 'm' is attached to a spring oscillates every 5 second. If the mass is increased by a 5 kg, the period increases by 3 second. Find its initial mass 'm'
Md Reply
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Practice Key Terms 4

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