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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

The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
Opeyemi Reply
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
it is the work done in moving a charge to a point from infinity against electric field
Ashok Reply
what is the weight of the earth in space
peterpaul Reply
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Thats why it can't have a constant value of g ....
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
please why is the first law of thermodynamics greater than the second
Ifeoma Reply
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
Mateshwar Reply
explain the lack of symmetry in the field of the parallel capacitor
Phoebe Reply
pls. explain the lack of symmetry in the field of the parallel capacitor
does your app come with video lessons?
Ahmed Reply
What is vector
Ajibola Reply
Vector is a quantity having a direction as well as magnitude
tell me about charging and discharging of capacitors
Ahemen Reply
a big and a small metal spheres are connected by a wire, which of this has the maximum electric potential on the surface.
Bundi Reply
3 capacitors 2nf,3nf,4nf are connected in parallel... what is the equivalent capacitance...and what is the potential difference across each capacitor if the EMF is 500v
Prince Reply
equivalent capacitance is 9nf nd pd across each capacitor is 500v
four effect of heat on substances
Prince Reply
why we can find a electric mirror image only in a infinite conducting....why not in finite conducting plate..?
Rima Reply
because you can't fit the boundary conditions.
what is the dimensions for VISCOUNSITY (U)
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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