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(17)
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Notice that the potential
is a scalar quantity. Instead of three (or more) equations to solve, there
is only one. It is this fact that makes the potential easier to work with
in most calculations. Also, just as with potential energy, the potential
of a field is not an exact quantity. Instead, it only has physical meaning
when changes in the potential are considered. In fact, the field potential
is related to the potential energy in exactly the same way the field is
related to the force (up to a change in sign), i.e., by
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(18)
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where q0 is the charge associated with the field under consideration.
Example:
Find the equipotential surfaces
for two charges of equal magnitude but opposite signs separated by a distance
a.

When dealing with electromagnetism, it is common to denote the electric potential by the symbol, V, which has units of volts. In terms of more fundamental units, a volt is defined as
Since electromagnetism is
a long range force, it is common to take the reference point, Va,
to be located at infinity, so that Va = 0. When this
choice is made, then DW
is just the work required to move a test charge in from infinity to the
point in question. Notice though, that there is no requirement that Va
be at infinity; occasionally, the inherent symmetry of the problem will
dictate another choice for the location of Va. An example
of this will be given later.
Let us now consider a single
point charge. What is the potential due to this charge? Recall that the
electric field for the charge is given by
Then we find, for a point charge
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(19)
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If we have a collection of discrete charges.
then the potential for the system is just the sum of the individual potentials:
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(20)
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Example:
What is the potential of
an electric dipole?

Assume that the two charges have a magnitude of q and are separated by a distance 2a. For simplicity, place the center of the dipole at the origin. Then for a point at a distance r from the origin and at an angle q from the dipole direction, we have

where
and
In the limit r >> 2a, this reduces
to
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(21)
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where p = 2aq is the electric dipole moment. Notice that the dipole is strongest along it's axis of orientation (q = 0, p) for any radius.
The electric dipole moment is found in many molecules and atoms. It is formed when the distribution of the charge is not symmetric about the physical distribution of the object. In these cases, there is an excess of positive charge at one side, and negative charge at the other. When the object is given a dipole moment by an external electric field, then the dipole is known as an induced electric dipole moment. When the external field is removed, the induced dipole moment will also disappear.
Example:
What is the potential of
a spherical shell of radius R with a charge Q?
For a spherical charge density, the electric field is given by
Since we are working with a shell instead of a solid sphere, this is split into three sections: r > R, r = R, and r < R. Let us look at each one separately.
Case 1: r > R.
In this case, the electric
field is given by
Since this is the electric field of a point charge, we have already found the potential. It is
Case 2: r < R.
In this case, the electric
field is zero (all of the charge is on the surface of the shell). So we
see that
where the constant is yet to be determined.
Case 3: r = R.
This case is what is called
a boundary condition. Since V is a continuous function,
the values of V(r>R) and V(r<R)
must be equal at r = R. Looking at V(r>R),
we see that, at r = R, we have
Thus, we see that the unknown constant for case 2 is just
Summing all this up, we have that, for a spherical
shell, the potential is given by
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(22)
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In this example we can see an interesting phenomenon. We can write Q as a density times a volume. As the volume is decreased, the density will increase. This causes a buildup of large potentials and fields at sharp points and corners. If the charge buildup is large enough, then the electric field generated by the charge will cause the air around it to ionize. This effect is known as a corona discharge and the best known example of it is lightning during a thunderstorm.
Example:
What is the potential of
an infinite line charge?
The change in the voltage is given by
The electric field of an infinite line charge
is given by
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(23)
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where l is the charge per unit length. Taking Dl = Dr r and summing over all the Dr, we get
where Va is our reference potential.
Recall that earlier I said that it is common to take the reference potential,
Va, to be located at infinity, so that Va
= 0. In this case, if we take r = ¥,
instead of getting Va = 0 (as we would want) we would
find Va = ¥.
Thus to have Va = 0, we must take some arbitrary distance
r = a. Then the potential becomes
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(24)
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Example:
How much energy is required
to move an electron through a potential difference of 1 V?
Using the relation between energy and electric potential we have
This combination shows up so frequently in particle physics that it has been given it's own designation, the electron volt, or eV. We frequently talk about particles with kinetic energies in the thousands, millions, billions or even trillion eV, so these are denoted keV, MeV, GeV and TeV respectively.
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(25)
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The SI unit of capacitance is the farad. From (25) it can be seen that one farad is equal to a coulomb per volt. Notice that, although the capacitance is defined in terms of the potential, the potential will always be proportional to the charge on the conductor, so the capacitance will depend only on the shape of the conductor.
Example:
For a spherical shell of
radius R with a charge Q, we saw that the potential is given
by
Therefore, the capacitance of a spherical shell
is
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(26)
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A capacitor is a system
of two conductors carrying equal and opposite charge. Note that they do
not have to have the same shape, but must have the same charge. In terms
of field lines, this means that all the field lines that originate on the
positive conductor must end on the negative one. Thus, the overall charge
of a capacitor is zero. Recall that the electric potential is defined
only up to an additive constant. This allows us to define the potential
of one of the conductors to be zero. Then, the potential that is used
in (25) is the potential difference between the two conductors.
In practice, the capacitance
is not difficult to calculate once the potential difference is known. The
problems arise in trying to calculate the potential difference for complex
arrangements of the conductors. As we shall see, even for simple arrangements,
this calculation depends upon the shapes of the various conductors and
the distances between them.
Example:
Consider two parallel conducting
plates, each with a surface area of A, separated by a distance d.
Let one plate carry a charge q and the other plate have a charge
-q. This is known as a parallel plate capacitor.

To find the capacitance we
have to find the electric field. For an infinite sheet of charge, the electric
field is given by
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(27)
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and points from the positive plate towards the
negative one. Here s
is the charge per unit area. For a finite plate, (if we neglect the effect
of the edges) we can replace s
by q/A. Notice that this electric field exists only in between
the two plates; outside them the field is zero. This can be seen by considering
the field lines. Only inside the plates do the field lines point in the
same direction. Once we have the electric field, we can find the potential
difference between the plates. It is
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(28)
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where d is the distance between the two
plates. Finally, once we have the potential, we can find the capacitance
from (25). Thus,
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(29)
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Notice that the capacitance is independent of charge, proportional to the size of the conductors and inversely proportional to the distance between them.
where V is the voltage difference between the two pieces of the capacitor. Since the voltage is related to the potential energy by
we can calculate the energy stored by replacing the change in electric potential with the average voltage established between the capacitor plates:
Thus, the energy stored in the capacitor is just
or, upon using the fact that q = CV,
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(30)
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Forming the ratio Q/V, we see that
the effective capacitance is given by
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(31)
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Thus, we see that we can replace two capacitors
in parallel by a single capacitor with an effective capacitance equal to
the sum of the capacitance of the replaced capacitors. This result holds
in general: We can always replace a chain of parallel capacitors by
a single capacitor whose capacitance is the sum of the replaced capacitance.
Symbolically, we write
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(32)
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If we have two capacitors hooked up in series, then the potential at the bottom of the first capacitor must equal the potential at the top of the second one, so the total potential difference across the two capacitors is the sum of the potential differences across each. The charge on each capacitor will be the same though. If the potential difference of the first capacitor is V1 and the potential difference of the second capacitor is V2 then

Forming the ratio 1/Ceff =
V/Q, we see that the effective capacitance is given by
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(33)
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The effective capacitance
of two capacitors in series is less than that of either one of the individual
capacitors. In general we have, We can always replace a chain of series
capacitors with a capacitor for which the reciprocal of the effective capacitance
is the sum of the reciprocals of the individual capacitors. Symbolically,
we write
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(34)
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(35)
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Since the original charge Q0
is still on the plates, the new capacitance is
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(36)
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Since the potential difference between the plates
of a parallel plate capacitor is just the electric field between the plates
times the separation d, the effect of the dielectric is to decrease
the electric field by the factor c.
If E0 is the original field without the dielectric,
the new field E is
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(37)
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Thus, the effect of a dielectric is to replace
the original potential, capacitance and electric field with (35), (36)
and (37) in all equations.
Another way of determining
the dielectric is to look at another quantity, the permittivity,
e. The permittivity
is related to the dielectric constant by the formula
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(38)
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