Equipotential Lines = Contours of constant V
E field
points downhill
Downhill is
always
perpendicular to
level
Conductors at
rest are
equipotential
Electric potential is a scalar.
Electric fields are vectors.
BOTH follow principle of superposition… with multiple
sources you can add the (field, potential) from each
source.
Rank the magnitudes (smallest to
largest) of the electric field at
point P in the three arrangements
shown.
A] all are the same
B] I, II, III
C] III, II, I
D] II, I, III
Rank the electric potentials at
point P (smallest to largest).
Where is on the x axis is the potential = 0?
(other than at infinity)
Take V = 0 at infinity.
Or choose E: potential is zero nowhere except infinity.
Ans. E
Where is on the x axis is the E field = 0?
(other than at infinity)
Or choose E: field is zero nowhere except infinity.
Ans. C ratio of distances = 2:1
Where is on the x axis is the potential = 0?
(other than at infinity)
Take V = 0 at infinity.
Or choose E: at more than one point shown.
Ans. E (both A & B have V=0)
Where is on the x axis is the E field = 0?
(other than at infinity)
Or choose E: off the left edge of the slide
Ans E. 15 units to the left of the -1 charge, the ratio of
distances is 2:1
Potential from an infinite plane
Potential from an infinite line
(From infinite sources, we can’t take V=0 at infinity.
But we are only ever interested in potential differences,
so it doesn’t really matter.)
E (none)
Which graph shows the potential from an infinite line of
negative charge density, where x = distance from line?
What if you have two sources, say an infinite line and a point
charge?
In that case, there is no “zero convention”… no preferred place
where we can all agree to make V=0. So it no longer makes
any sense to speak of THE potential at a point. We can still find
the potential difference between two points, however.
What is the potential difference between a point a distance r to the left of
the point charge, and a point immediately below the point charge on the
plane? Recall that the field from an infinite plane is E=
finding E from V
If V = -4x + 4y2 (x,y in meters, V in volts) , what is the E field at the
origin?
A] 0
B] 4 V/m in the +y direction
C] 4 V/m in the -y direction
D] 4 V/m in the +x direction
E] 4 V/m in the -x direction
If V = -4x + 4y2 (x,y in meters, V in volts) , where is the E field = 0?
A] at the origin
B] at x=1, y=1
C] at x=1, y=-1
D] at both B and C
E] nowhere.
Two conductors in proximity form a “capacitor”: they have a
capacity to hold a charge Q (+Q on one and -Q on the other)
with a voltage difference V.
C=Q/V
Note: Capacitance C is italicized. Coulomb C is not.
Capacitance is a property of the geometry of the conductors
(and depends on whether there is vacuum or material
between them, as we will see later.)
This is because of linearity (superposition.) If I double the
source charges, I will double the field, and double the
potential.
“Reservoir” analogy
The charge on a capacitor (±Q) is like the water
The potential is like the depth of the water
The capacitance is like the area of the lake
So Q=CV
If I put ±2 microcoulombs of charge on the sphere and plate,
there is a potential difference of 2 V.
What is the capacitance?
A] 0
B] Cannot determine, since not symmetric
C] 1 microfarad = 10-6 C/V
D] 106 farads
If I put ±2 microcoulombs of charge on the sphere and plate,
there is a potential difference of 2 V.
If I put ±4 microcoulombs of charge on the sphere and plate, what
will be the potential difference?
A] 0
B] 2 V
C] 4 V
D] 8 V
To calculate the capacitance of a pair of conductors, put
+Q on one and -Q on the other, and calculate the
potential difference.
(Sometimes we refer to the capacitance of an isolated
conductor. In that case, we put charge Q on the
conductor and ask what is the potential (compared to
infinity.)
Let’s do a cylindrical capacitor example.
Parallel plate capacitor
Two infinite parallel sheets carry charge densities ± 
What is the electric field at point 1?
Two infinite parallel sheets carry charge densities ± 
What is the electric field at point 2?
The potential difference between the plates is d
Capacitors “in parallel” (= same potential across them.)
Ceq = C1 + C2
Capacitors “in series” (= same charge on each)
1/Ceq = 1/C1 + 1/C2
What happens to the potential difference if I keep the same
Charge on the plates, but increase separation?
A] V increases
B] V decreases
C] V is unchanged
V increases. Where does this energy come from?
Answer: it takes work to move the plates apart.
Where is this energy stored??
It is stored in the field between the plates.
The field is the same, but the volume of space is bigger.
The “energy density” is the same, but there is more energy.
How much energy?
It is proposed to store solar energy by filling a water
tank during the day, raising the water from ground level.
(At night, the water drains turning a generator.)
A tank is 20 m high and has vertical sides. When half
full, it stores 1 million Joules. How much energy will it
store when full?
A] 1 million Joules
B] 2 million Joules
C] 4 million Joules
D] 20 million Joules
“Dielectric” = a fancy word for a real insulator.
In real insulators, charges can move, but only very small
distances… the electrons can shift a bit with respect to
the nucleii, but they can’t hop from one nucleus to the
next.
What happens to the electric field in a capacitor if we
add a dielectric material?
Let’s sketch on the board, and then I’ll quiz you.
By superposition, the field in the dielectric is the sum of the
original capacitor field, plus the field from the polarization
surface charge.
The total field is therefore
A] bigger than the original capacitor field
B] smaller than the original capacitor field
C] the same as the original capacitor field
Answer: smaller than the original field.
We can write that the new field is the original field, divided by
K, the “dielectric constant”
• What is the new potential across the capacitor?
A] same as the old potential, without the dielectric
B] bigger than the old potential, by a factor of K
C] smaller than the old potential, by a factor of K
If I put a charge ±Q on a parallel plate capacitor and insert a
dielectric, what happens to V?
A] it increases
B] it decreases
C] it stays the same
If I put a charge ±Q on a parallel plate capacitor and insert a
dielectric, what happens to V?
The field in the dielectric is reduced by a factor of K. So the potential goes down.
What happens to the potential energy?
A] it goes up by a factor of K
B] it goes up by a factor of K2
C] it goes down by a factor of K
D] it goes down by a factor of K2
E] it stays the same.
If I put a charge ±Q on a parallel plate capacitor and insert a
dielectric, what happens to V?
What happens to the potential energy?
It may be counterintuitive, but the potential energy goes down by a factor
of K (not K2).
U=Q2/(2C); the capacitance goes up by a factor of K.
(or U = CV2/2 … the voltage drops by K, the capacitance increases by K)
Note: the energy density for a given E field in a dielectric is u 
1
2
E 2
The field goes down by a factor of K, but epsilon adds a factor of K.

If the potential energy goes down, where does the energy go?
• The field pulls the dielectric into the capacitor, giving it
kinetic energy. (Very small, here.)
• Application: Optical tweezers
With a fixed charge on the plates, the dielectric is pulled into the
capacitor. The field and field energy go down.
What happens if, instead, the capacitor is held at a fixed
potential?
The energy goes up! U = CV2/2, and the capacitance goes up.
But we cannot conclude that you must push the dielectric in (in
fact, you don’t have to) … the extra energy comes from the
battery.