Chapter 19
Electric Potential Energy
and the
Electric Potential
19.5 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
19.5 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
19.5 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
19.5 Capacitors and Dielectrics
19.5 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
Example. How large would the plates have to be for a 1 F air-filled
parallel plate capacitor with plate separation a) d=1 mm, b) d=1 µm ?
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!
19.5 Capacitors and Dielectrics
Conceptual Example 11 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
19.5 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:
Energy = 1/2 qV = 1/2 CV2 = q2/(2C) --> useful forms to solve problems
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?
Energy = 1/2 CV2 = 1/2(10)(3)2 = 45 J
This is the same amount of energy stored in a 4.6 kg mass
suspended 1 m above the ground.
Chapter 20
Electric Circuits
20.1 Electromotive Force and Current
In an electric circuit, an energy source and an energy consuming device
are connected by conducting wires through which electric charges move.
20.1 Electromotive Force and Current
Within a battery, a chemical reaction occurs that transfers electrons from
one terminal to another terminal.
The maximum potential difference across the terminals is called the
electromotive force (emf).
The emf has units of Volts, e.g. a 12 V battery has an emf of 12 V
20.1 Electromotive Force and Current
The electric current is the amount of charge per unit time that passes
through a surface that is perpendicular to the motion of the charges.
Δq
I=
Δt
One coulomb per second equals one ampere (A).
20.1 Electromotive Force and Current
If the charges move around the circuit in the same direction at all times,
the current is said to be direct current (dc) --> e.g. simple circuits with
batteries are normally dc.
If the charges move first one way and then the opposite way, the current is
said to be alternating current (ac) --> e.g. the current coming out of your
electric outlet is ac.
20.1 Electromotive Force and Current
Example 1 A Pocket Calculator
The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hour
of operation, (a) how much charge flows in the circuit and (b) how much energy
does the battery deliver to the calculator circuit?
(a)
(
)
Δq = I (Δt ) = 0.17 ×10 −3 A (3600 s ) = 0.61 C
ΔV = -W/q
(b)
Energy = Charge ×
Energy
= (0.61 C )(3.0 V ) = 1.8 J
Charge
20.1 Electromotive Force and Current
Conventional current is the hypothetical flow of positive charges that would
have the same effect in the circuit as the movement of negative charges that
actually does occur.
20.2 Ohm’s
Law
It is found experimentally that the
current flowing through a circuit is
proportional to the voltage across the
circuit, i.e.,
I is proportional to V
The resistance (R) is defined as the
ratio of the voltage V applied across
a piece of material to the current I through
the material.
20.2 Ohm’s
Law
OHM’S LAW
The ratio V/I is a constant, where V is the
voltage applied across a piece of material
and I is the current through the material:
V
= R = constant
I
or
Resistance
SI Unit of Resistance:
volt/ampere (V/A) = ohm (Ω)
V = IR
20.2 Ohm’s
Law
Resistance, R, represents to what extent
the current can flow freely in the circuit, i.e.
the larger R, the more the electrons scatter
with atoms in the material.
These scatterings slow down electrons and
transfer energy as heat to the material.
To the extent that a wire or an electrical
device offers resistance to electrical flow,
it is called a resistor.
20.2 Ohm’s
Law
Example 2 A Flashlight
The filament in a light bulb is a resistor in the form
of a thin piece of wire. The wire becomes hot enough
to emit light because of the current in it. The flashlight
uses two 1.5-V batteries to provide a current of
0.40 A in the filament. Determine the resistance of
the glowing filament.
V
3.0 V
R= =
= 7.5 Ω
I 0.40 A