Chapter 17
Electric Potential
The Electric Potential Difference Created by Point Charges
When two or more charges are present, the potential due to all of the charges
is obtained by adding together the individual potentials.
Example: The Total Electric Potential
At locations A and B, find the total electric potential.
The Electric Potential Difference Created by Point Charges
VA
(8.99 ×10
=
VB
9
)(
) (
)(
)
N ⋅ m 2 C 2 + 8.0 ×10 −-98 C
8.99 ×109 N ⋅ m 2 C 2 − 8.0 ×10 −-98 C
+
= +240 V
0.20 m
0.60 m
(8.99 ×10
=
9
)(
) (
)(
)
N ⋅ m 2 C 2 + 8.0 ×10 −-98 C
8.99 ×109 N ⋅ m 2 C 2 − 8.0 ×10-9−8 C
+
=0V
0.40 m
0.40 m
Equipotential Surfaces and Their Relation to the Electric Field
An equipotential surface is a surface on
which the electric potential is the same everywhere.
kq
V=
r
Equipotential surfaces for
a point charge (concentric
spheres centered on the
charge)
The net electric force does no work on a charge as
it moves on an equipotential surface.
Equipotential Surfaces and Their Relation to the Electric Field
The electric field created by any charge
or group of charges is everywhere
perpendicular to the associated
equipotential surfaces and points in
the direction of decreasing potential.
In equilibrium, conductors are always equipotential surfaces.
Equipotential Surfaces and Their Relation to the Electric Field
Lines of force and equipotential surfaces for a charge dipole.
Equipotential Surfaces and Their Relation to the Electric Field
The electric field can be shown to
be related to the electric potential.
If you move a distance Δs and
the electric potential changes
by an amount ΔV, the component
of the electric field in the
direction of Δs is the negative
of the ratio of ΔV to Δs,
or more compactly,
ΔV
E=−
Δs
ΔV/Δs is called the potential gradient.
Equipotential Surfaces and Their Relation to the Electric Field
Example: The Electric Field and Potential
Are Related
The plates of the capacitor are separated by
a distance of 0.032 m, and the potential difference
between them is VB-VA=-64V. Between the
two equipotential surfaces shown in color, there
is a potential difference of -3.0V. Find the spacing
between the two colored surfaces.
Equipotential Surfaces and Their Relation to the Electric Field
E =−
Δs = −
ΔV
−64 V
=−
= 2.0 ×10 3 V m
Δs
0.032 m
ΔV
−3.0 V
−3
=−
=
1.5
×10
m
3
E
2.0 ×10 V m
The gradient equation can be simplified for a parallel plate capacitor
with voltage V across it and with plates separated by distance d as:
V
E=
d
Capacitors and Dielectrics
A parallel plate capacitor consists of two
metal plates, one carrying charge +q and
the other carrying charge –q.
It is common to fill the region between
the plates with an electrically insulating
substance called a dielectric.
Advantages of dielectrics:
•  they break down (spark) less easily than air
•  the capacitor plates can be put closer together
•  they increase the ability to store charge
Capacitors and Dielectrics
THE RELATION BETWEEN CHARGE AND POTENTIAL
DIFFERENCE FOR A CAPACITOR
The magnitude of the charge in each place of the
capacitor is directly proportional to the magnitude
of the potential difference between the plates.
q = CV
The capacitance C is the proportionality constant.
SI Unit of Capacitance: coulomb/volt = farad (F)
The larger C is, the more charge the
capacitor can hold for a given V
E0
Capacitors and Dielectrics
THE DIELECTRIC CONSTANT
If a dielectric is inserted between the plates of
a capacitor, the capacitance can increase markedly.
Dielectric constant
Eo
κ=
E
Since E ≤ E0 --> κ ≥ 1
Some lines of force now
terminate or start on
induced surface charges
E ≤ E0
E0
E
Capacitors and Dielectrics
Capacitors and Dielectrics
THE CAPACITANCE OF A PARALLEL PLATE CAPACITOR
Eo = q (ε o A)
Eo V
E=
=
κ
d
& κε A #
q = $ o !V
% d "
Parallel plate capacitor
filled with a dielectric
<-->
q = CV
κε o A
C=
d
C increases for --> increasing κ, increasing A, decreasing d
Capacitors and Dielectrics
Example: A Computer Keyboard
One common kind of computer keyboard is based on the
idea of capacitance. Each key is mounted on one end
of a plunger, the other end being attached to a movable
metal plate. The movable plate and the fixed plate
form a capacitor. When the key is pressed, the
capacitance increases. The change in capacitance is
detected, thereby recognizing the key which has
been pressed.
The separation between the plates is 5.00 mm, but is
reduced to 0.150 mm when a key is pressed. The
plate area is 9.50x10-5m2 and the capacitor is filled with
a material whose dielectric constant is 3.50.
Determine the change in capacitance detected by the
computer.
Capacitors and Dielectrics
C=
C=
(
3.50 )(8.85 ×10 −12 C2
=
(N ⋅ m ))(9.50 ×10
(
3.50 )(8.85 ×10 −12 C2
=
(N ⋅ m ))(9.50 ×10
κε o A
d
κε o A
d
2
−5
m2
-3
0.150 ×10 m
2
-3
5.00 ×10 m
ΔC = 19.0 ×10 −12 F
−5
m2
) = 19.6 ×10
−12
) = 0.589 ×10
F
−12
F
Example. How large would the plates have to be for a 1 F parallel plate
capacitor with plate separation a) d=1 mm in air, b) d=1 µm separated
by paper?
a)  d = 1 mm
C = κε0A/d = ε0A/d (since κ = 1 for air)
--> A = Cd/ε0 = (1)(.001)/(8.85 x 10-12) = 1.13 x 108 m2
If the plates were square, they would be of length and width
(A)1/2 = (1.13 x 108)1/2 = 11 km --> very large!!
b) d = 1µm = 10-6 m ~ the thickness of paper – use paper: κ = 3.3
--> A = Cd/(κε0) = (1)(10-6)/(3.3 x 8.85 x 10-12) = 3.42 x 104 m2
and (A)1/2 = (3.42 x 104)1/2 = 185 m --> about 2 football fields in length,
improving!
Capacitors and Dielectrics
Conceptual Example: The Effect of a Dielectric When a
Capacitor Has a Constant Charge
An empty capacitor is connected to a battery and charged up.
The capacitor is then disconnected from the battery, and a slab
of dielectric material is inserted between the plates. Does the
voltage across the plates increase, remain the same, or
decrease?
Effect of a Dielectric When a Capacitor Has a Constant Charge -- continued
Before inserting dielectric:
q = C 0 V0 ,
C0 = ε0A/d
After inserting dielectric:
q = C V,
C = κε0A/d = κC0
C0 V0 = q = C V = κC0 V
--> V = V0/κ --> V < V0 for κ > 1
i.e. the voltage across the plates
decreases when the dielectric is
inserted if charge is held constant
Capacitors and Dielectrics
ENERGY STORAGE IN A CAPACITOR
Capacitors store charge at some potential difference, therefore they store EPE,
since by definition for a point charge q,
EPE = qV .
Your book shows that the energy stored in a capacitor can be written several ways:
2
1
1
q
Energy = qV = CV 2 =
2
2
2C
Using Energy = 1/2 CV2, C = κε0A/d, and V = Ed , the energy of a
capacitor can be written as,
Volume = Ad
& κε o A #
2
Energy = $
!(Ed )
% d "
1
2
We can then calculate the energy density of a capacitor as,
Energy density =
Energy
Volume
= κε o E
1
2
2
(valid in general for the
energy density stored
in an electric field)
Example. How much energy is stored in a 10 F capacitor
with a potential difference of 3 V across it?
1
1
2
2
Energy = CV = (10 ) (3) = 45J
2
2
This is the same amount of energy stored in a 4.6 kg mass
suspended 1 m above the ground.