Chapter 17
LECTURE NOTES
Electric Potential, Electric Energy, Capacitance
Consider a small positive point test charge q in a constant electric field E.
E
q
+
A
F
d
B
The charge feels a force in the direction of E: F = qE (recall the definition of E). This electric
force does work W = F ۰ d in moving the charge from A to B (displacement d). We have W =
qEd. This results in a loss of potential energy for the charge  PE = -W = -qEd.
We now define a quantity independent of the size of the test charge: 0
The electric potential difference V is the change in potential energy per unit charge: V =
PE
= -Ed. We see V has units J/C which we call a volt: 1V = 1 J/C. We see E has
q
magnitude –V/d and can be written with units V/m (which we show in homework to be
equivalent to N/C). The electric field is the negative of the potential gradient.
Note: V is a scalar, not a vector. It can be +, 0, or – but it has no direction to bother with, so
potential problems are often simpler than electric field problems.
Note: A common unit of electric energy is the electron volt. Using  PE = qV we define 1eV
as the energy gained when an electric charge e = 1.6 x 10-19 C is accelerated through a potential
difference of 1 volt.
Note: Potential difference (V = VB – VA), measured in volts (V), is often referred to as voltage.
Note: We often write V or say potential or voltage when we mean V (potential difference).
The gravitational potential energy of a body is mgh when it is h above some arbitrary zero level.
The gravitational potential would be the potential energy difference between the two points per
unit mass. The gravitational field points down (from high to low potential) and is the direction a
mass naturally falls.
The electric potential in this example was higher at A than at B. The field points from high to
low potential. This is the direction a positive charge naturally goes.
kQ
is the potential (in
r
volts) a distance r (in m) from the center of a charge Q (in C), where k = 9 x 109 Nm2/C2.
It is convenient for a point charge to define V = 0 at r =  . Then V =
Chapter 17
LECTURE NOTES
Note: If Q is +, V is +; if Q is -, V is -.
We can represent the electric potential graphically by connecting points of equal potential with
curves or surfaces. No work is done when charges move along equipotential surfaces. These
equipotential surfaces are always perpendicular to the electric field. E field lines are thus
perpendicular to these surfaces and point from high to low regions of potential. Note the surface
of a conductor is an equipotential surface. There can be no  V inside a conductor in the static
case (or charges would move) so V is constant inside the conductor and equal to the value on its
surface.
Note: The potential energy of a pair of charges Q1 and Q2 can be written PE = Q2V1 =
kQ1Q 2
r
when the charges are a distance r apart.
y
A positive charge Q1 (4.00C) is located
at (0, 0.400m). A negative charge
Q2 (-3.00  C) is located at (0.300m, 0).
Find the electric potential at the origin due
these charges.
+
Example
x

2

9  10 9 Nm2 (4  10 6 C)
kQ1
C

V1 =
= 90,000V
r1
(.400m)


9  10 9 Nm2 (3  10 6 C)
kQ 2
C

V2 =
= –90,000V
r2
(3.00m)
Thus, V = V1 + V2 = 0V at the origin.
2
Capacitance
A capacitor is a device used to store electric charge. The simplest capacitor takes the form of a
pair of parallel plates of area A separated by a distance d. This gap may be filled with air or
some other (dielectric) material.
If we connect a battery to such a parallel plate capacitor, positive charge can be said to flow
away from the positive terminal of the battery to one plate of the capacitor which accumulates a
charge +Q. At the same time, positive charge can be said to be withdrawn from the other plate
(leaving behind charge –Q) to flow into the negative terminal of the battery. The battery does
work (and loses energy) driving these charges. When the charge on the capacitor creates its own
potential difference equal but opposite to that of the battery charge no longer flows and we are
back in the static case studied in this chapter. Moving charge (currents) will be studied later.
Chapter 17
LECTURE NOTES
It is found that the charge Q (in coulombs) on one plate is proportional to the potential difference
across the plates (equal to the voltage of the charging battery in our example).
Thus Q = constant ۰  V or Q = VC where C represents the proportionally constant and we have
written V for  V.
The capacitance of the device is C = Q/V in units of coulombs/volt; we write 1 farad (F) = 1C/V.
Note: Capacitance (C) is measured in farads (F)
Charge (Q) is measured in coulombs (C)
Potential (V) is measured in volts (V).
The above relation may be more easily remembered as Q = VC by those who have seen this
“Shopping Channel” on TV.
It can be shown that for a parallel plate capacitor with an air gap C =
capacitance in farads,  0 is a constant = 8.85 x 10
-12
0A
where C is the
d
C2
2
Nm 2 , A is the area of one plate (in m ) and
d is the separation of the plates (in m).
Just as 1 C is a lot of charge, a 1F capacitor has a large capacitance. Most devices used in
electronics are in the pF to  F range.
Often an insulating sheet (or dielectric) is placed between the plates of a capacitor instead of air.
This has the effect of increasing the capacitance to C =  A/d when    0 is the permitivity of
the material and  the dielectric constant (which is about 1 for air). A dielectric has polar
molecules of unequal charge distribution that orient themselves along the electric field between
the plates. This enforced charge separation lowers the potential difference between the plates
and since C = Q/V, the lower V increases C.
Dielectrics have additional advantages. They allow the plates to be closer together without risk
of touching and have a greater resistance than air to sparks jumping the gap.
Electric Energy
A charged capacitor stores energy in the electric field between the plates. The energy U (in
1
1
1
joules) stored is: U = QV  CV 2  Q 2 / C. Further, the energy per unit volume or energy
2
2
2
1
density u =  0 E 2 . This last relation is true not just in a capacitor but wherever an electric field
2
is present.
Chapter 17
LECTURE NOTES
Example: Two sheets of aluminum foil (each 1.00m by .300m) are separated by a piece of
newspaper (0.1mm thick) forming a parallel plate capacitor. The device is connected to a 9-volt
battery. Calculate the capacitance, charged stored, and energy stored for this device. Take the
dielectric constant of newspaper to be 5.
12 C
A 5(8.85  10 Nm2 )(1m)(.3m)
Then, C =
=130nF

d
1  10 4 m
Q = VC = (9V)(130۰10-9F) = 1.2  C
1
1
U = QV  (1.2 ۰ 10-6C)(9V) = 5.4  J.
2
2
2
Question: How would you measure the thickness of a piece of newspaper?