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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
Viola Reply
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Jyoti Reply
what are questions that are likely to come out during exam
King Reply
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Jyoti Reply
watt is electricity.
Adam
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Jyoti
If a point charge is released from rest in a uniform electric field will it follow a field line? Will it do so if the electric field is not uniform?
Sadaqat Reply
Maxwell's stress tensor is
Ami Reply
Yes
doris
neither vector nor scalar
Anil
if 6.0×10^13 electrons are placed on a metal sphere of charge 9.0micro Coulombs, what is the net charge on the sphere
Rita Reply
18.51micro Coulombs
ASHOK
Is it possible to find the magnetic field of a circular loop at the centre by using ampere's law?
Rb Reply
Is it possible to find the magnetic field of a circular loop at it's centre?
Rb Reply
yes
Brother
The density of a gas of relative molecular mass 28 at a certain temperature is 0.90 K kgmcube.The root mean square speed of the gas molecules at that temperature is 602ms.Assuming that the rate of diffusion of a gas in inversely proportional to the square root of its density,calculate the density of
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Gifty
Under which topic
doris
what is electrostatics
Yakub Reply
Study of charges which are at rest
himanshu
Explain Kinematics
Glory Reply
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?
Nya
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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Samuel Reply
Practice Key Terms 4

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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