Physics 1302/1402
Lecture 11
Capacitance
Previously…
• We finished our discussion of electric potential V
• In particular, we studied the potential difference
between two charged parallel plates:
V  El
Today…
• We will introduce the capacitor – an electronic
device that stores electrical energy.
Converting work into potential energy
• We must do work to assemble a distribution of
charges.
• Since the electrostatic force is a conservative force,
this work turns into stored electric potential energy.
Converting work into potential energy
• To assemble the three charges shown, we must do work.
• The work to bring q1 in from infinitely
far away is
W1  0
• The work to bring q2 in from infinitely
far away in the field due to q1 is
kq1q2
W2 
a
• The work to bring q1 in from infinitely
far away in the field due to q1 and q2 is
kq1q3 kq2 q3
W3 

a
a
Converting work into potential energy
• Total work is
WT  W1  W2  W3
kq1q2 kq1q3 kq2 q3



a
a
a
• This is the total potential energy stored
in this arrangement of charges.
EXAMPLE 1
• If the three particles shown have identical charge q
and mass m, and if they are simultaneously released
from their positions on the triangle, what will be their
speed v when they are far away from their initial
positions?
Storage of electrical potential energy
• Batteries: stored energy is released via chemical
reactions that produce a potential difference.
•
•
•
•
large energy reservoir in a small volume
constant voltage over lifetime of discharging
low currents
used in direct-current (DC) circuits
• Capacitors: energy is stored directly in an electric field,
released by moving charges around.
•
•
•
•
capable of high current, so high power
rapid re-charging
voltage decreases with discharging
generally used in alternating current (AC) circuits
Capacitors
• Simplest example of a capacitor is the
parallel-plate capacitor
• Two parallel conducting plates
separated by a small gap
Capacitors
• Initially the plates are neutral, so there is no electric field.
• We than charge the plates, so that one plate has charge +Q and
the other has charge –Q.
• This produces an electric field between the plates.
• If the gap between the plates is
small, the field between the plates
is uniform and points
perpendicular to the plates.
• Outside the plates, the field is ~ 0.
Capacitor voltage
• Surface charge density on each plate is   Q / A
• This produces an electric field between the plates of
E  4k   /  0  Q /  0 A
• Recall: permittivity of free space is 𝜀0 = 1/(4𝜋k) =
8.85×10–12 C2/Nm2
• We showed previously that the potential difference V
between the plates is V  Ed
so
V  Qd /  0 A
Capacitance: definition
V  Qd /  0 A
• Rearranging,
Q   A 0 / d V
 CV
• C  Q / V  A 0 / d is called the capacitance of our
parallel plate capacitor.
• C depends only on the geometry of the plates, and of
the material in between them.
Capacitance: units
• Capacitance: C  Q / V
• C is the charge stored on the capacitor plates per volt
of potential difference across the plates.
• Units = [C/V] = Farads
• Typical capacitors have capacitance in the pF
(picofarad) – mF (milifarad) range.
Capacitors: examples
• Numbers given are:
capacitance,
(maximum) operating
voltage
• e.g., 4700𝜇F 35V
• Sometimes
capacitance is given
by numbers like
103K, 224K – these
are, e.g.,
• 103 = 10 + 000 (3
zeroes) = 10,000 pF
• 224 = 22 + 0000 (4
zeroes) = 220,000 pF
• Capacitors store electrical energy:
 https://www.youtube.com/watch?v=EoWMF3VkI6U
EXAMPLE 2
• A capacitor consists of square conducting plates 25
cm on a side and 5.0 mm apart, carrying charges ±1.1
μC. Find (a) the electric field inside the capacitor, (b)
the potential difference between the plates, and (c)
the stored energy.
Energy stored in a capacitor
• Let’s move a small amount of charge dQ from one
plate to the other, when the potential difference
between the plates is V.
• Potential difference is work per unit charge, so
dW  VdQ
• Changing the charge on the plates changes the
potential difference. Since Q = CV,
dQ  CdV
dW  VdQ
dQ  CdV
So…
dW  CVdV
• As the charge on the plates increases, V increases.
This means that the work required per unit charge
increases as well.
dW  CVdV
• Let’s start with uncharged plates, so V = 0 initially.
• The work required to charge the plates to a potential
difference is V is then
V
W   dW   CVdV
0
1
 CV 2
2
• This work is stored in the capacitor as potential energy.
EXAMPLE 2,
cont’d
• A capacitor consists of square conducting plates 25
cm on a side and 5.0 mm apart, carrying charges ±1.1
μC. Find (a) the electric field inside the capacitor, (b)
the potential difference between the plates, and (c)
the stored energy.
EXAMPLE 3
• An uncharged capacitor has parallel plates 5.0 cm on
a side, spaced 1.2 mm apart. (a) How much work is
required to transfer 7.2 μC from one plate to the
other? (b) How much work is required to transfer
another 7.2 μC?
Dielectrics
• So far we have assumed that the gap between the
plates of our capacitor is filled with air.
• Real capacitors usually have a thin layer of insulating
material – a dielectric – sandwiched between the
plates.
• Dielectrics contain electric dipoles, but (almost) no
free charges.
Dielectrics
• Dipoles orient in the field of
the capacitor in such a way
that the field inside the
capacitor is decreased.
• This decreases the potential
difference for a given charge.
• Increases the charge you can
store at a given V.
• Using a dielectric increases the
capacitance C.
Dielectrics
• Dielectrics are characterized by a dielectric constant κ.
• Electric field in the capacitor is reduced to

E
 0
• Capacitance of a parallel-plate capacitor filled with dielectric is
C 
0 A
d
• Higher κ = higher C = more energy storage for a given V.
Working voltage of a capacitor
• If the electric field between the capacitor plates is too
high, the dielectric will break down
• This sets the “working voltage” of the capacitor –
maximum safe potential difference between the
plates
EXAMPLE 4
• Let’s do example 2 again, but with a dielectric in the
gap between our plates:
• A capacitor consists of square conducting plates 25
cm on a side and separated by a 5.0 mm thick sheet
of glass (κ = 5.6). The plates carry charges ±1.1 μC.
Find (a) the electric field inside the capacitor, (b) the
potential difference between the plates, and (c) the
stored energy.
Pre-class readings for next class:
• Y&F, Chapter 24, Sections 24.1–24.2
Practice problems:
•
Y&F, Chapter 24, problems 24.3, 24.23, 24.25