Equipotentials
These are lines through space
which define regions of constant
voltage or electric potential.
The first rule to note with
equipotentials, is that the surface of
a conductor is always one
equipotential. It has to be, because
if there was a difference in
potential along the surface, then
charges would move through the
conductor until the difference in
potential was eliminated.
Conductor
Equipotenial
Flux Line
Conducting Surface
Shown is a parallel plate capacitor with the
electric flux lines drawn in. If we were
to draw the equipotential lines (lines
along which the electric potential is
constant) on this diagram, what lines
should we draw?
a) Horizontal lines, parallel to the flux lines
b) Vertical lines, parallel to the plates
c) Diagonal lines
d) Concentric circles, centered on the
midpoint between the plates
+Q
-Q
Correct Answer : B
The lines are parallel to the plates. The plates
are conductors, so the surfaces of the plates
themselves are equipotentials, so it is not
surprising that all of the equipotentials are
parallel to the plates in a parallel-plate
capacitor. This illustrates another rule, that
equipotentials run perpendicular to field
lines, which in this case are horizontal. Since
the electric field is constant inside the
capacitor, the distance between each pair
equipotential is the same.
Energy stored in a Capacitor
Besides being a device which stores charge, a capacitor also
stores energy. We can say the electric field inside the capacitor
contains energy and it would be interesting to know how much
energy this amounts to. It will be useful to remember that a
capacitor has the following properties
Q = total charge on each plate (+ on one plate, - on the other)
V = voltage difference between the two plates = E d
E is the electric field strength in the region between the plates
d is the distance between the plates
A is the area of the plates
C = capacitance of the plates = Q/V = e A/d = k eo A/d
The unit of Capacitance is the Farad
named after our old friend Michael Faraday
1 Farad is the capacitance of a capacitor which will put 1
Coulomb of charge on its plate for each 1 Volt of potential
difference between the plates.
Most capacitors are measured in microFarads or picoFarads.
A 1 F capacitor is unusual, expensive and can be dangerous!
You’ll notice that checking one’s results by units is not as easy
in electromagnetism as it was in mechanics. One always has
constants like k and e with funny units which are hard to
remember. Nevertheless if you are working the problem and
have a book this is still an excellent way to check your answer.
Suppose we allow one of the positive charges
on the left plate to fly across to the right
hand plate where the negative charges are.
What will happen?
a) The charge on the capacitor and the energy
stored in it will both be reduced
b) The charge and the energy in the capacitor
will both increase
+q
c) The charge will decrease on one plate and
increase on the other, and the energy will be
unchanged
d) The charge will decrease but the energy
stored will increase
+Q
-Q
Correct Answer – A
The charge on both plates is reduced, because the positive
plate lost one of its charges, and one of the negative charges
on the other plate has had its field effectively “canceled out”
by the arrival of a new positive charge.
Since the charge on both plates is reduced, it follows that the
strength of the electric field between them is also reduced, and
it seems logical that a weaker electric field has less energy
stored in it.
We can check this by looking at the work done. The force on
the charge generated by the field was pointing in the direction
it moved, so the work done by the field on the charge was
positive, and this means that the charge gained energy (in the
form of kinetic energy) and the field lost energy. When the
charge hits the plate on the other side the kinetic energy is
dissipated as heat. So the field loses energy permanently.
What amount of energy is lost by the field
when the charge flies across from one
plate to the other? In other words, what
work is done on the charge by the field in
moving it across?
a) W = E Q A
b) W = Q V d
+q
c) W = Q q d
d) W = E q d
Here W = work done, E = electric field
strength, Q = charge on plates, q = charge
of the charge being moved across, A =
area of plates, d = distance between plates
+Q
-Q
Correct Answer – D
Remember that the work done by a force F moving an object a
distance d is W = F d. But we also know that the force exerted
by an electric field of strength E on a charge q is given by F =
E q.
Therefore the work done by the field on the charge is
W=Fd=Eqd
Note this q is the charge being moved across, not the total
charge on the plates, Q.
How does the work done on the charge q
moving across from one plate to the other,
depend on the voltage difference between
the plates. If W = E q d then which of the
following is true?
a) W = V Q
b) W = V q d
+q
c) W = V q
d) W = V
Correct Answer – C
Recall that the relationship between electric field and
potential difference or voltage is simple
E = V/d
The electric field is the voltage per unit distance, so
V=Ed
Therefore W = E q d = V q
The best way to figure out the total energy stored in
the field in the capacitor would be to calculate
how much work would need to be done to move
every positive charge on the left plate across to
the right plate. After that the total charge on the
plates would be zero, and the electric field and
voltage would both be zero. All the energy
would be gone. So after we had moved half of
the charges across, what could we say about the
work needed to move the next charge across
compared to the work needed to move the first
charge across?
+q
a) It would be just the same
b) It would be half of the work on the first charge
+Q
-Q
c) It would be twice the work on the first charge
Correct answer – B
Once you have moved half the charges across from one plate to
another, the total charge on the plates is now half what it was
to begin with, which means the voltage between the plates is
also halved, since V = Q/C (remember that C is constant).
Since the electric field is only half what it was before, this
means the force on the charge is half what is was before and
therefore the work done is also halved, since
W=Fd=Eqd
If d and q are unchanged and E is halved, then so is W.
So if we think of draining the energy away slowly by moving the
charge across, the change in energy DU when you decrease the
charge on the plates by an amount Dq is DU = W = V Dq. This means
the total energy on the plates, U, is the sum of of all these losses of
energy, which means it is the sum of V Dq, which is to say, the area
under the curve on the plot of V versus Q. What is this result?
V
a) U = ½ V Q
b) U = V Q
c) U = 2 V Q
d) U = V Q2
U = area
under line
Q
Correct Answer – A
The area under the curve is easy in this case, since the curve is a
straight line from Voltage = V, charge = Q (the initial state of the
capacitor) to Voltage = 0, charge = 0 (the final state). The area
under the line is half the area of a rectangle with one side of
V
length V and the other side of length Q.
The area of the whole square = Q V, so
the area of half the square is
(V,Q)
U = ½ Q V. Other ways of writing this
are U = ½ C V2 and U = ½ Q2/C,
Since C = Q/V. If you can produce a
given voltage with a battery, than the
higher the capacitance the more energy
your capacitor can store.
U = area
under line
(0,0)
Q
What does this tell us about the amount of energy which can be
stored in an electric field?
If U = ½ C V2, then since V = E d we have
U = ½ C E2 d2
Now we also know that C = e A /d so
U = ½ e A E2 d
Now if A is the area of the plates and d is the distance between
them, than A d = V the volume of the space between the plates,
which is to say, the volume of the space that our electric field
occupies. So
U/V = energy density in the electric field = ½ e E2
This result applies for any electric field, not just inside a capacitor