Electric Potential
As for all basic forces of nature, the electrostatic force is a conservative force. Therefore
the work done by the electrostatic force is independent of the path taken and we can
define an electrostatic potential energy that can help us to implement energy conservation
in mechanical problems.
U (r ) = − ∫ F(r ') ⋅ dr '
r
r0
In words, the electrostatic potential energy equals the work done by the electrostatic force
in bringing the particle from the location of interest (r) to the reference location (r0).
Knowing the potential energy allows us to derive the electrostatic force. We simply
calculate the gradient of U(r) to obtain:
F(r ) = −
∂U
∂r
The gradient is a vector that points in the direction of steepest ascent.
Electric Potential Energy for Point Charge: As an example lets calculate the electric
potential energy associated with a point charge q in the vicinity of another point charge
Q. Symmetry tells us that the potential can only depend on the distance between the two
charged objects. We choose infinite separation to be the point of reference. Then at a
separation, r the electric potential is
U (r ) = − ∫
Qq
Qq
dr =
2
4πε 0 r
∞ 4πε 0 r
r
We notice that the electrical potential energy is equivalent to the gravitational potential
energy though with masses substituted by charges. We check that we recover Coulombs
law when calculate the gradient of the potential energy:
F(r ) = −
∂  Qq 
Qq

 =
rˆ
∂r  4πε 0 r  4πε 0 r 2
Definition of the Electric Potential
Just as we defined an electric field from the electrostatic force, we define an electric
potential from the electric potential energy by dividing out a "test charge". The idea is to
associate an electric potential with a charge distribution. We imagine measuring this
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potential by measuring the electric potential energy for a test charge in the vicinity of this
charge distribution, then divide this function with the magnitude of the test charge to
obtain a function that characterizes the charge distribution.
U (r )
q
V (r ) =
We see that the units for the electrical potential is J/C. We use this unit so often in
electromagnetism that we have given it a special name Volt or in short V. Using this new
unit we see that the unit for electric field N/C can also be written as V/m.
Calculating the Electric Potential
A charge distribution yields a spatially varying electric potential. We can calculate the
potential from a general charge distribution starting from the expression for the electric
potential associated with a point charge. This is trivially derived from the expression for
the electrical potential energy associated with a point charge by dividing with the test
charge:
V (r ) =
Q
4πε 0 r
Note that a positive point charge creates a positive potential. The expression looks a lot
like the expression for the electric field from a point charge but there are crucial
differences that could be overlooked at first glance:
•
•
The electric potential is a scalar the electric field is a vector
r enters to the first power in the denominator for the electrical potential. It enters as
r2 in the denominator of the expression for the electric field from a point charge.
The superposition principle allows generalization from this simple result. If we have
several discrete point charges, the electrical potential is calculated via:
V (r ) = ∑
i
Qi
4πε 0 ri
If we have a continuous charge distribution, the electric potential is derived through a
suitable integration:
V (r ) =
ρ (r ')d 3 r
1
∫
4πε 0 V '
R
σ (r ')d 2 r
1
V (r ) =
∫
4πε 0 S '
R
V (r ) =
λ (r ')dr
1
∫
4πε 0 L ' R
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for volume charge distribution
for surface charge distribution
for line charge distribution
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You will notice the similarity of these expressions with those we used to calculate the
electric field form continuous charge distributions. On close inspection, you will notice
that the expressions for V(r) are significantly simpler than those for E(r) because the
former is a scalar while the latter is a vector. This is part of the reason that we care to
introduce and work with the electrical potential.
Deriving V from E
If we have E(r) and we want V(r), we use the following expression that follows from the
way we derive a potential energy from a force:
V (r ) = − ∫ E(r ) ⋅ dr
r
r0
The integration goes from the point of reference (r0) where the electric potential by
definition is zero to the point (r) where we want to calculate the electric potential. As an
example, we calculate the potential difference between
Two charged metallic plates:
Assume the charge on the left and right plates is +Q and -Q respectively. The field
outside the plates must vanish (an application of Gauss law will show this). Therefore the
charge density on the inner surfaces of the plates is σ=Q/A and from Gauss law, we easily
derive that the electric field between the plates is
E(r ) =
σ
zˆ
ε0
where the unit vector points from the positively charged plate to the negatively charged
plate. The electric potential between the plates is
V (r ) = − ∫ E(r ) ⋅ dr = V (r0 ) −
r
r0
σ
z
ε0
−σ
+σ
The total electric potential difference between the plates is
∆V = V+ − V− =
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σ
d
ε0
z
d
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V is Constant in a Metal
In electrostatics, there can be no electric field in the interior of a conductor. If there were,
then charge would flow and we would not yet have achieved the electrostatic condition.
We used this result to derive the fact that charge lies on the surface of a conductor from
Gauss law. From this result comes also an interesting result about the electric potential.
Because
V (r ) = − ∫ E(r ) ⋅ dr
r
r0
We can say the any point within a volume of space where the electric field vanishes has
the same electric potential. Think about it for a while and this result should not be a
surprise at all. If the electric field vanishes there can be no force on any charged particle
which in turn means that there can be no electrical potential energy difference and hence
no electric potential difference.
This result makes it clear why the interior of a metallic vehicle is relatively safe from
lightning in a thunderstorm. Certainly if the car formed a complete closed shell (no
windows) then the shell and the interior and the passengers would be part of the same
equipotential surface and hence there would be no danger within the vehicle from electric
field differences. Even if we allow for windows it turns out that the vehicle offers relative
safety as long as the windows are small compared to a characteristic electromagnetic
wavelength. Once we go to an open vehicle, it cannot be guaranteed that the passengers
will be safe on an equipotential surface. If we go to a fiberglass Corvette the vehicle
offers no safety at all.
We can illustrate the ability of a metal to shield from electric fields by enclosing a small
radio in a metallic cage. The radio signal is transmitted by setting up charge
distributions on large antennae that in turn generate an electric potential that moves
charge around on the antenna of the radio. This charge motion is detected by the
electronics within the radio and converted to motion of the loudspeaker membrane and
sound waves that we detect with our ears. However, if we enclose the radio in a metal
cage then it becomes part of an equipotential volume. The charge on the antenna no
longer experiences an electric force and the radio is no longer affected by the electric
fields being generated by the antennae. The reason that the signal is shielded although
there are openings in the wire net, is that the openings are smaller than the wavelength for
the electromagnetic waves. This is a point that we will return to later.
Deriving E from V
Just as we can derive a force from a potential energy, we can derive the electric field
vector from the electric potential:
F(r ) = −
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∂U
∂r
⇒ E(r ) = −
Page 4
∂V
∂r
02/07/004
As an example of an application of this, we calculate the electric field around conducting
spheres held at the same potential but have different radius. We know that a charged
sphere is indistinguishable from a point charge at the center of the sphere throughout the
region of space outside the sphere. Thus, the electric potential outside a sphere must also
be the same as that around a point charge:
V (r ) =
Q
4πε 0 r
If we charge two spheres to electric potential Vs then the charge on the surface of the
sphere is evidently given by
Q = 4πε 0 RV
Where R is the radius of the sphere. The larger sphere has the larger charge. This seems
sensible. However, if we instead look at the charge density on the surface of the sphere
we find:
4πε 0 RV ε 0V
Q
σ =
=
=
2
R
4πR
4πR 2
The expression implies that the surface charge density is greater on the smaller sphere.
Thus, it should not be a surprise that the electric field is also largest around the smaller
sphere:
E(r ) = −
V
∂
∂ Q
Q
V (r ) = −
=
rˆ = s rˆ
2
∂r
∂r 4πε 0 r 4πε 0 r
r
Specifically we see that the magnitude of the electric field on the surface of spheres held
at the same potential is inversely proportional to the radius of the spheres.
We check this result by a few demonstration experiments:
1. Draw sparks between spheres charged to the same potential using the Wimshurst
machine. Notice that the spark occurs between the smaller radius spheres. This is the
experimental indication that the electric field is greater around the spheres with
smaller radius.
2. We image the electric field pattern around a charged ovoid using the Wimshurst
machine and fibers in an oil solution. The fibers concentrate around the region with
small radius of curvature indicating again that the field is largest there.
3. Charge an ovoid with the van de Graff generator. Connect a metallic object to the
large and small curvature part of the ovoid and see that the charge collected is greater
when we connect to the part of the ovoid that has a smaller radius of curvature.
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