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Lectures on Chapter 20: Electric Potential
and Capacitance
Gravitational force and electrostatic force are similar:
Recall that we can define a gravitational potential energy. If the
work done to
lift an object from
to
is
then the potential energy gained by the object is
For example, a mass m a distance h above the ground has potential energy
=1.75 true in
mgh.eps
The energy is measured with respect to the ground
,
but
we could put the zero of energy anywhere. It would just add a constant to
Ua and Ub but wouldn't change the potential energy
difference U(h) - U(ground).
Since gravity is a conservative force,
Ub - Ua is independent
of the path taken between a and b. All that matters is the
endpoints.
In analogy with this, we can define the change in the
electric potential energy.
If we have a charge q that finds itself in an electric field
,
then it feels a force
is due to other charges, e.g., another point charge. The electric
potential energy associated with dragging the charge from a to its present
position b is the negative of the work done in getting it there.
goes along the path we took from a to b. Like gravity,
is a conservative force so that
depends only on the
endpoints a and b, not on the path between them.
Notice that the sign of
will change in
.
So if the field comes from positive charges and q>0, we do work in
dragging q toward the charges and
increases. If q<0,
then q is attracted to the positive charges and drags us toward
them;
decreases and is negative. (It's like the
difference between uphill and downhill.)
Let's take q>0. Notice that the bigger q is, the more work we do in
dragging it around, and the bigger the potential energy is. Remember
how it was
convenient to divide
by q to get .
is independent of q which feels it but doesn't produce it, e.g.,
consider the force produced by a point charge q1.
depends on q and q1, but
depends only on q1.
Similarly, it is convenient to divide U by q to get something
independent of q.
V is called the electric potential.
This depends on only on the charges that produce .
Just as was the force felt by a unit charge (q=1),
is the potential energy
of a unit charge, i.e.,
is the potential energy per unit charge.
Units
= joules (after all, it's an energy)
So electric potential and electric potential energy are different.
Sometimes the electric potential is referred to as voltage.
Notice that if there is potential difference
, the
potential energy gained by moving a proton with charge +e through this
difference (``uphill" or against the field) is
1 electron volt.
=2.0 true in
1eV.eps
An eV is a unit of energy:
In general, moving a charge q through a voltage drop changes its
potential energy by .
So a big voltage can give a small if q is small. For example,
rubbing your hair with a balloon can charge the balloon to several thousand
volts but the amount of charge is small
C) so is
small
you won't get electrocuted.
The gravitational analog is
. Just as with
gravity,
depends only on where a and b are, not
on the path between them. Thus it doesn't matter whether you take path 1 or
path 2, is the same. (The gravitational analog of
was F/m = g = acceleration.)
=2.0 true in
voltagePaths.eps
You can think of
as gravitational potential (if
you multiply by m, you get gravitational potential energy). Think of
different distances up a hill. Higher points on the hill have greater
gravitational potential. Going uphill is like acquiring a higher voltage by
dragging a positive charge opposite to the field. ``Danger: High Voltage" are
like ``Danger: Falling rocks" signs.
Equipotential Surfaces
An equipotential surface is the set of points which all have the same electric
potential or voltage. These points are usually a surface.
If a and b are on the same equipotential surface, then
Vb = Va and
. Look at
How can
? One way:
everywhere along (like inside a conductor). Another way:
is a line integral along the path parameterized by
.
adds up all the components of
along .
.
If
, then
,
. So
on
surfaces
.
=2.0 true in
Eds.eps
Examples:
=5.0 true in
equipotentials.eps
All points on the surface have the same potential but different surfaces have
different potentials. Gravitational analogy: Every point at the same altitude
h has the same gravitational potential gh.
to
equipotential surfaces.
points along the fall
line on a slope.
=3.5 true in
hill.eps
Bigger
Bigger V (Swim upstream)
Lots of charge
Charges set up . They also set up V.
Note is different from U.
Potential in a Uniform Field
Electric potential is the analog of height on a hill. Notice that
the height difference between points a and b is the same
as the height difference between point b and c.
=2.0 true in
hill2.eps
Similarly the potential difference between points a and b is the same
as the potential difference between points b and c in the figure of
equipotential planes in a uniform electric field. Dragging a mass
uphill corresponds to dragging a positive charge from a to c or
from a to b.
=2.0 true in
equipotUniformE.eps
The potential difference between points a and c is given by
We can take out of the integral because the field is
constant. It's a uniform field that is the same everywhere. So
we see that in a uniform field the potential is E times the
perpendicular distance between equipotential surfaces.
Dimensional analysis: Notice that [V]=[Ed] implies that
voltage or potential has the same units as electric field times
length or (Newtons/Coulomb) meters = Joules/Coulomb=Volts.
Similarly, [E]=[V/d] which means that electric field has units
of Volts/meter = V/m. Earlier we learned that
[E]=[Force/Charge]=Newtons/Coulomb. So 1 V/m = 1 N/C.
Potential of a Point Charge
Consider a point charge q > 0 at the origin. Infinitely far away, a test
charge q0 feels no force
at .
So let's set
and
because there is
no potential energy and no potential
at
. What is the potential V(r) a distance r away
from q? If we start at r and push q0 > 0 to infinity, then
This is less than 0 because we lose potential energy; we are going downhill.
=2.5 true in
VofPtCharge.eps
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-V(r) |
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V(r) |
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(7) |
If
,
.
So V changes sign if q sign.
V(r) falls off slower than
. Notice that V(r) >0 which
means that if we started at
and dragged a charge q0 to the
point r, its potential energy would increase:
One of the great benefits of V is that it is a scalar, so we don't have to
find the components of a vector. V(r) is a scalar field, i.e., a number is
associated with every point in space.
Superposition
To find the potential of a group of charges at a point , add the
potential r at due to each charge:
Because V is a scalar, we can add magnitudes. We don't have to worry about
vector components.
Potential of an Electric Dipole
=2.5 true in
dipolePot.eps
For r>>d,
and
Azimuthal symmetry (rotational symmetry about z-axis)
no dependence.
In the equational plane,
because V+ and V- cancel
(V+ = - V-).
Continuous Charge Distribution
Given
, find
.
Recipe for Finding the Potential due to a Continuous Charge
Distribution
Method I
- 1.
- Divide charge distribution into pieces with charge dq.
- 2.
- dq produces a potential
where r is the distance from dq to the point P.
- 3.
-
.
Integrate over the charge distribution. (Notice this
assumes V=0 at
Method II
- 1.
- Use Gauss' Law to find
if there is symmetry.
- 2.
- Use
Example Using Method II: Charged Planar Sheet
Consider an infinite (insulating) sheet with uniform positive charge density
. What is the change in potential in going from z1 to z2,
where z1 and z2 are distances from the sheet?
=3.0 true in
planarCharge.eps
Solution
Use
- 1.
- Find
using Gauss' Law:
Notice that
is uniform, i.e., it's the same everywhere.
- 2.
- We saw earlier that the potential difference between 2 points
in a uniform field is Ed. Since z2 is further from the
sheet of charge than z1, we expect
V(z2) - V(z1)<0.
So
V(z2) - V(z1)=-Ed. We can also go through the math again:
V(z2) - V(z1) |
= |
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V(z2)-V(z1) |
= |
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Calculating the field from the potential
We've seen how to calculate the potential V if we know
.
How do we calculate if we know V?
Suppose
( is in the direction.)
Then
If
, then
If
, then
In general we can write
where is called ``the gradient of V.''
What does this mean? Recall the equipotential surfaces, each with
constant V. is perpendicular to the surfaces and points from
high potential to low potential, i.e., ``downhill". So if you put
a + charge where V=5 volts, it
would feel a force in the direction of V=0 volts.
.
Think of a hill. Equipotential corresponds to equi-altitude. would
point straight downhill, i.e. along the ``fall line".
Electric Potential Energy due to a System of Charges
The electric potential energy of a system of fixed point charges is
equal to the
work done to assemble the changes, bringing one at a time in from infinity.
=2.0 true in
threePtCharges.eps
Consider assembling 3 point charges. Bring in one charge at a time.
No work to bring in q1, but q1 sets up a potential
where r is the distance from q1. Now
bring in q2. The energy we store in doing this is
where r12 is the distance between q2 and q1.
(The work we do is minus the work done by the field, so
.)
Note that the general formula for the potential energy of a pair of point
charges a distance r apart is
q2 also sets up a potential
When we bring in q3, we add more potential energy.
So the total electric potential energy stored is
An Isolated Conductor
Recall that
inside an isolated
conductor. If were not zero, then the free conduction electrons
would feel
and they would run. A charged, isolated
conductor has charges on the surface because the charges get as far apart
as possible. They arrange themselves such that
inside.
There is no charge inside because Gauss' law tells us that
V = Constant throughout a conductor, i.e., a conductor is an
equipotential. Pick 2 points a and b in the conductor or on its
surface.
=2.0 true in
conductorE.eps
surface of a conductor
If has a component along
the surface, the charges would run along the surface. Here can be
either due to charges on the surface of the conductor or due to
an external field
. If an uncharged conductor is
placed in an external field, the conduction electrons feel the field and
race to the surface, arranging themselves until they cancel the field
inside the conductor. This ``induced charge" produces
a field
such that
inside the
conductor.
=2.0 true in
conductorEind.eps
Capacitors
So far we've been considering a single isolated conductor. Now consider 2
isolated conductors. This is a capacitor. The conductors are
called capacitor plates. Capacitors are important circuit elements.
They are also used to store energy.
To understand how they work, consider a parallel plate capacitor
consisting of 2 parallel conducting plates a distance d apart. Each
plate has area A. When the capacitor is charged, its plates have equal
and opposite charges +q and -q. Each plate has a constant potential,
but there is a potential difference V between the plates.
(Conventional notation:
potential difference between the
plates.) V is proportional to q. (``q" is called ``the charge of
the capacitor," even though both plates are charged.)
=2.0 true in
parallelplate.eps
The constant of proportionality C is called the capacitance of the
capacitor.
Units
1 farad is huge. Typical units:
and
1p F =
10-12F. Notice that there is an electric field between the capacitor
plates. points from the +q plate to the -qplate.
=2.0 true in
capacitorE.eps
The +q plate is at a higher voltage than the
-q plate. The potential difference between the plates is
Notice that
. How do we get
from this? Recall Gauss' law:
This implies
. Since
,
. Thus
or
.
Calculating Capacitance
C depends only on the geometry of the conductor, i.e., their size,
shape, and separation. C is independent of q and . In a typical
problem, you are given the geometry of the conductors and asked to find
the capacitance C. (C > 0 always.)
Recipe to find the capacitance
- 1)
- Assume +q on one plate and -q on the
other plate (or surface charge density
on one plate,
on another plate).
- 2)
- Use Gauss' law to find
between plates.
- 3)
- Use
to find
the potential difference between the plates. Note that we want
.
- 4)
- Use
to find C.
(The charge you put on the plates in step 1 will cancel out.)
Example: Parallel Plate Capacitor
Find the capacitance of a parallel plate capacitor with area A and
separation d.
neglect fringing
fields.)
=2.0 true in
parallelplate.eps
Solution: Put +q on the top plate and -q on the bottom plate.
Then the top plate has
surface charge density and
the bottom plate has
. The charge on each plate
is attracted to the side facing the other plate. No excess charge is
on the outer surfaces. Find using Gauss' law.
=3.0 true in
parallelplateE.eps
where we used
Find using
where we used
and
.
Find
Notice that this just depends on the size and separation of the plates,
not on q or . C increases with A and decreases with
increasing d. It is more convenient to express
(Before we used different units:
C2/(N-m2)).
If the plates are 1 mm apart
(d = 10-3m) and C = 1F, how big are
the plates?
If the plates are square, then each side is
Example: Cylindrical Capacitor
Consider 2 coaxial cylinders of length L and radii a and b. Find
the capacitance C.
=2.0 true in
cylindricalcap.eps
Solution: Assume the inner cylinder has
charge +q and
. The
outer cylinder has -q and
.
Find using Gauss' law. The Gaussian
surface is a cylinder of length
.
Find :
where we switched limits so that
.
Find C:
Again C only depends on geometrical quantities.
Spherical Capacitor
A spherical capacitor consists of 2 concentric spherical conducting shells
of radii a and b. Find the capacitance C.
=2.0 true in
sphericalcap.eps
Solution: Assume the inner sphere has charge +q. Find between
the spheres. We know that outside the sphere of charge +q, is the
same as for a point charge:
We could also use Gauss' law:
Find the potential difference between the spheres
Suppose we have an isolated conductor of radius a.
=1.0 true in
oneSphere.eps
We can assign it a
capacitance by assuming the other (outer) sphere is at infinity. So if we
set
, then
Numbers for Van de Graaf
cm = 0.2 m
pF/m
Energy released
If this is released in 1 second, power
Watts.
Storing Energy in the field
We have seen that capacitors store charge. It turns out that
charging capacitors is also a way to store energy.
=2.0 true in
pumpCharge2.eps
To charge up a capacitor,
you have to remove electrons from the positive plate and carry them to the
negative plate. In doing so, you fight against the electric field, which
is pulling them back toward the positive plate and repelling them from the
negative plate. How much work does this take? If a voltage has
already built up,
. The work done in transferring
charge dq (imagine dq > 0 being transferred from the negatively charged
plate to the positively charged plate) is
The total work in going from q = 0 to q = Q is
This work is stored as potential energy in the capacitor
Since
. So
The bigger the voltage or charge, the more the energy that is stored. An
example of energy stored in a capacitor is a camera flash.
Where is the energy residing? It is viewed as being stored in the
electric field . It is customary to define an energy
density u as the potential energy per unit volume (of the space with
E). Consider a parallel plate capacitor with voltage difference
and a uniform electric field
.
Using
, we can write
Although we derived this result for the special case of a parallel plate
capacitor, it holds in general. (Just imagine the field being uniform in
a tiny element of space.) This formula is true even if there isn't a
capacitor. If ,
. Note that
superposition does not hold:
So add fields and then square to get u.
Capacitor with Dielectric
It is common to find capacitors that are 1F but aren't 10 km on a side.
How do we increase C without increasing the dimensions of the capacitor?
Answer: We can place an insulator, such as plastic, glass, or oil,
between the capacitor plates. This insulator is called a dielectric. It increases the capacitance by a numerical factor called the dielectric constant. Thus, if we place a dielectric with
dielectric constant between the plates of a capacitor, the
capacitance increases:
where C0 is the capacitance without the dielectric.
For a parallel place capacitor,
= 1 for vacuum.
for all other materials. So
, i.e., the capacitance increases.
Why does this happen? Microscopically the molecules in the dielectric
become polarized by the electric field between the plates of a charged
capacitor. ``Polarized" means that the electric dipole moments of the
molecules line up in the field. These dipole moments are either
``permanent" dipole moments (like H20) or ``induced" dipole moments (or
both, i.e., permanent moments enhanced by induction). An induced dipole
moment occurs when the external field ``stretches" a molecule by
separating the positive and negative charges a little. This creates
=1.25 true in
dipoleMolecule.eps
Inside the dielectric there is no excess charge. On the
surface near the positively charged plate, the dielectric has a build up
of negative charge. Similarly, the negatively charged plate attracts a
buildup of positive charge in the dielectric. The dielectric is neutral
overall, i.e., no net charge, but it is polarized. The dielectric charges
``screen" the charges on the plate, making them less repulsive to each
other and more willing to stay.
=2.0 true in
dielectricMolecules.eps
For example, the
positive charge on the upper plate is attracted to the negative charge on
the surface of the dielectric, so it is more willing to stay than before
the dielectric was there. (This is like feeling more comfortable at a
party if you know some people there.) Thus the dielectric enables us to
put more charge on the capacitor plates, i.e., to increase the
capacitance.
=2.0 true in
dielectric.eps
Notice that the induced field
in the dielectric
is opposite to
. As a result the total field
between the plates is reduced.
For a given amount of (free) charge Q on the plates, is
reduced when the dielectric is present. (``Free'' charge refers
to the charge that is not induced.)
where Q is the free charge and
.
To achieve a given voltage between the plates
Therefore a capacitor can hold more free charge with the dielectric.
Charging the capacitor to voltage stores energy U
More charge must be pumped onto the dielectric capacitor to achieve a
given voltage . This takes more work. So more energy
is
stored.
Putting free charge Q on the capacitor stores energy U
It takes less work to put Q on the dielectric capacitor because the
charge is screened.
Gauss' law when a dielectric is present:
Capacitors in Circuits
Capacitors are important circuit elements. One way to charge a capacitor
is to attach it to a battery. You can think of a battery as a pump which
takes plus charge from the negatively charged plate and adds it to the
positively charged plate.
=2.0 true in
pumpCharge.eps
The battery keeps doing this until the voltage
across the terminals of the battery equals the voltage across the
capacitor plates. Let's draw a circuit diagram to show the setup for
doing this. An electric circuit is the path through which current
flows.
=3.0 true in
firstCircuit.eps
Current starts to flow and the capacitor starts charging once the switch
S is closed. Once C is charged, it will stay charged even if S is
opened, i.e., +q and -q stay on the plates.
Sometimes we have more than one capacitor in a circuit. It is convenient
to replace combinations of capacitors with an equivalent capacitor
Ceq. This
simplifies the circuit. There are two basic combinations.
Capacitors in Parallel
=2.5 true in
Cparallel.eps
Capacitors are connected in parallel when a potential difference applied
across their combination results in that potential difference being
applied across each capacitor. The equivalent capacitor has the same
amount of charge as all the parallel capacitors combined and the same
potential drop as all capacitors combined.
In the case shown
Ceq is the equivalent capacitance of the single capacitor.
=2.0 true in
Ceq.eps
In general,
Add the capacitances for capacitors in parallel.
Capacitors in Series
=2.0 true in
Cseries.eps
Capacitors are connected in series when the potential difference applied
across the combination is the sum of the potential differences across each
capacitor. Ceq has the same q and the same as the whole
combination.
=2.0 true in
CseriesPart.eps
Notice that each capacitor in the series has the same amount of charge on
it. To see this, note that the boxed conductor is electrically isolated and
therefore has no net charge. The charges are merely separated into +qand -q. No charge can be transferred to the isolated element.
By the definition of being in series,
=2.0 true in
Ceq.eps
In general, for n capacitors in series
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Clare Yu
2002-02-14