Lecture 7
Potential from a distribution of charges
1
V =
4⇡✏0
X qi
§ Smooth distribution
1
V =
4⇡✏0
X qi
i
ri
1
=
ri
4⇡✏0
Z
⇢
dV
r
i
§ Calculating the electric potential from a group
of point charges is usually much simpler than
calculating the electric field
• It’s a scalar
Electric Potential from Two Oppositely
Charged Point Charges
§ The electric field lines from two oppositely charge point charges are a little
more complicated
§ The electric field lines originate on the positive charge and terminate on the
negative charge
§ The equipotential lines are always perpendicular to the electric field lines
§ The red lines represent positive
electric potential
§ The blue lines represent negative
electric potential
§ Close to each charge, the
equipotential lines resemble
those from a point
charge
22
Electric Potential from Two Identical Point
Charges
§ The electric field lines from two identical point charges are
also complicated
§ The electric field lines originate on the positive charge and
terminate at infinity
§ Again, the equipotential
lines are always
perpendicular to
the electric field lines
§ There are only positive
potentials
§ Close to each charge, the
equipotential lines
resemble those from
a point charge
23
Example: Superposition of Electric Potential
(1)
§ Assume we have a system of three
point charges:
q1 = +1.50 µC
q2 = +2.50 µC
q3 = -3.50 µC
§ q1 is located at (0,a)
q2 is located at (0,0)
q3 is located at (b,0)
a = 8.00 m and b = 6.00 m
Question: What is the electric potential
at point P located at (b,a)?
30
Example: Superposition of Electric Potential
(2)
r
Answer:
§ The electric potential at point P is
given by the sum of the electric
potential from the three charges
1
r2
r3
31
Conductor in E-field
§ Charges can move freely
along conductor boundary.
§ Recall: electrostatic
shielding: E=0 inside
§ Surface of a conductor is an
equipotential surface.
§ E-field is perpendicular to
conductor’s surface
7
Electric Potential Energy for a
System of Particles
§ So far, we have discussed the electric potential energy of a point
charge in a fixed electric field
§ Now we introduce the concept of the electric potential energy of a
system of point charges
§ In the case of a fixed electric field, the point charge itself did not
affect the electric field that did work on the charge
§ Now we consider a system of point charges that produce the electric
potential themselves
§ We begin with a system of charges that are infinitely far apart
• This is the reference state, U = 0
§ To bring these charges into proximity with each other, we must do
work on the charges, which changes the electric potential energy of
the system
37
Electric Potential Energy for a
Pair of Particles (1)
§ To illustrate the concept of the electric potential energy of a system of
particles we calculate the electric potential energy of a system of two
point charges, q1 and q2 .
§ We start our calculation with the two charges at infinity
§ We then bring in point charge q1
• Because there is no electric field and no corresponding electric force, this action
requires no work to be done on the charge
§ Keeping this charge (q1) stationary, we bring the second point charge (q2)
in from infinity to a distance r from q1
• That requires work q2V1(r)
38
Electric Potential Energy for a
Pair of Particles (2)
So, the electric potential energy of this two charge system is
where
Hence the electric potential of the two charge system is
If the two point charges have the same sign, then we must do positive work on the
particles to bring them together from infinity (i.e., we must put energy into the
system)
If the two charges have opposite signs, we must do negative work on the system to
bring them together from infinity (i.e., energy is released from the system)
39
Electric Potential Energy
Consider three point charges at fixed positions.
Question:
What is the electric potential energy U of the assembly of
these charges?
q1
d12
d13
q2
d23
q3
11
Example: Electric Potential Energy (2)
Answer:
§ The potential energy is equal to
the work we must do to assemble
the system, bringing in each
charge from an infinite distance
§ Let’s build the system by bringing
the charges in from infinity, one
at a time
41
Example: Electric Potential Energy (3)
§ Bringing in q1 doesn’t cost any work
§ With q1 in place, bring in q2
12
§ We then bring in q3.The work we must do to bring
q3 to its place relative to q1 and
✓q2 is then: ◆
U = U13 + U23 = kq3
Utot = U12 + U13 + U23 = k
✓
q1
q2
+
d13
d23
q1 q2
q1 q3
q2 q3
+
+
d12
d13
d23
◆
A sum of pair interaction energy.
42
Alternative way
Let’s calculate energy of a particle in a total potential
created by all other particles. And then sum over all
particles.
✓
Potential created by 2 &3 at
position of 1 is
Potential created by 1 &3 at
position of 2 is
Potential created by 1 &2 at
position of 3 is
V1 = k
✓
q2
q3
+
d12
d13
◆
◆
q1
q3
V2 = k
++
d12
d23
✓
◆
q1
q3
V3 = k
++
d13
d23
42
§ Total energy:
Utot = q1 V1 + q2 V2 + q3 V3 (?)
Utot = q1 V1 + q2 V2 + q3 V3 =
✓
◆
q2
q3
q1 k
+
+
???2???
d12
d13
✓
◆
I counted each interaction twice!
q1
q3
q2 k
+
+
d12
d23
q1 V1 + q2 V2 + q3 V3
◆
✓
Utot =
2
q3
q1
q3 k
+
1X
d13
d23
Utot =
q i Vi
✓
◆
2
N
q1 q2
q1 q3
q2 q3
= 2k
+
+
d12
d13
d23
Capacitors
§ Capacitors are devices that store energy in an electric
field.
§ Capacitors are used in many every-day applications
• Heart defibrillators
• Camera flash units
§ Capacitors are an essential
part of electronics.
• Capacitors can be
micro-sized on computer
chips or super-sized for
high power circuits such as
FM radio transmitters.
1
Capacitance
§ Capacitors come in a variety of
sizes and shapes.
The world's first
integrated circuit
included
one capacitor
Capacitor controlling power which enters Fermilab,
a particle accelerator laboratory in Batavia, IL
2
Capacitance
§ Concept: A capacitor consists of two
separated conductors, usually called plates,
even if these conductors are not simple
planes.
§ We will define a simple geometry and
generalize from there.
§ We will start with a capacitor consisting of
two parallel conducting plates, each with area
A separated by a distance d.
§ We assume that these plates are in vacuum
(air is very close to a vacuum).
3
Parallel Plate Capacitor (1)
+
+
+
-
-
+
+
+
- -
§ We charge the capacitor by placing
• a charge +q on the top plate
• a charge -q on the bottom plate
§ Because the plates are conductors, the charge will distribute itself
evenly over the surface of the conducting plates.
§ The electric potential, V, is proportional to the amount of charge on
the plates.
V = dE
E=
⇥
0
q
=
0A
dq
V =
0A
4
Parallel Plate Capacitor (2)
§ The proportionality constant between the charge q and the
electric potential difference V is the capacitance C.
§ We will call the electric potential difference V the
“potential” or the “voltage” across the plates.
§ The capacitance of a device depends on the area of the
plates and the distance between the plates, but does not
depend on the voltage across the plates or the charge on
the plates.
§ The capacitance of a device tells us how much charge is
required to produce a given voltage across the plates.
5
Definition of Capacitance
§ The definition of capacitance is
§ The units of capacitance are coulombs per volt.
§ The unit of capacitance has been given the name Farad
(abbreviated F) named after British physicist Michael Faraday
(1791-1867)
§ A farad is a very large capacitance
• Typically we deal with
µF (10-6 F),
nF (10-9 F),
or pF (10-12 F)
© Pixtal/age Fotostock RF
6
Charging/Discharging a Capacitor (1)
§ We can charge a capacitor by connecting the capacitor to a
battery or to a
DC power supply.
§ A battery or DC power supply
is designed to supply charge
at a given voltage.
§ When we connect a capacitor
to a battery, charge flows
from the battery until the capacitor is fully charged.
§ If we then disconnect the battery or power supply, the capacitor will
retain its charge and voltage.
§ A real-life capacitor will leak charge slowly, but here we will assume ideal
capacitors that hold their charge and voltage indefinitely.
7
Parallel Plate Capacitor (1)
§ Consider two parallel conducting plates separated by a distance d
§ This arrangement is called a parallel plate capacitor.
§ The upper plate has +q and the lower plate has –q.
§ The electric field between the plates points from the positively
charged plate to the negatively charged plate.
§ We will assume ideal parallel plate capacitors in which the electric
field is constant between the plates and zero elsewhere.
§ Real-life capacitors have fringe field near the edges.
9
Parallel Plate Capacitor (2)
We can calculate the
electric field between
the plates using Gauss’
Law
10
Parallel Plate Capacitor (3)
§ Now we calculate the electric
potential across the plates of the
capacitor in terms of the electric
field.
§ We define the electric potential
across the capacitor to be V.
§ We carry out the integral in the
direction of the blue arrow.
11
Parallel Plate Capacitor (4)
Remember the definition of capacitance…
… so the capacitance of a
parallel plate capacitor is
Variables:
A is the area of each plate
d is the distance between the plates
Note that the capacitance depends only on
the geometrical factors and not on the
amount of charge or the voltage across the
capacitor.
12
Example: Capacitance of a Parallel Plate
§ We have a parallel plate capacitor
constructed of two parallel plates,
each with area 625 cm2 separated by a
distance of 1.00 mm.
Question: What is the capacitance of
this parallel plate capacitor?
Answer: A parallel plate capacitor constructed out of square
conducting plates 25 cm x 25 cm separated by 1 mm has a
capacitance of about 0.5 nF.
13
Example: Capacitance, Charge, and
Electrons …
Question: A storage capacitor on a random access
memory (RAM) chip has a capacitance of 55 nF. If
the capacitor is charged to 5.3 V, how many excess
electrons are on the negative plate?
Answer:
Idea: We can find the number of excess electrons on
the negative plate if we know the total charge q on
the plate. Then, the number of electrons n=q/e,
where e is the electron charge in coulomb.
Second idea: The charge q of the plate is related to
the voltage V to which the capacitor is charged:
q=CV.
15
Capacitors can be dangerous!
§ 1 eV ~ 10000K
§ 100 eV ~ 100000K!
29
Cylindrical Capacitor (1)
§ Consider a capacitor constructed of two collinear
conducting cylinders of length L
• Example: coax cable
§ The inner cylinder has radius r1 and
the outer cylinder has radius r2
§ Both cylinders have charge per
unit length λ with the inner cylinder
having positive charge and the outer
cylinder having negative charge
16
Cylindrical Capacitor (2)
§ We apply Gauss’ Law to get the electric field between the two
cylinders using a Gaussian surface with radius r and length L as
illustrated by the red lines
§ … which we can rewrite to get an
expression for the electric field
between the two cylinders
r1< r < r2
17
Cylindrical Capacitor (3)
§ As we did for the parallel plate capacitor, we define the voltage
difference across the two cylinders to be V = |V1 – V2|
§ Thus, the capacitance of a cylindrical capacitor is
Note that C depends on
geometrical factors only.
18
Spherical Capacitor (1)
§ Consider a spherical capacitor formed by two concentric
conducting spheres with radii r1 and r2
19
Spherical Capacitor (2)
§ Let’s assume that the inner sphere has charge +q
and the outer sphere has charge –q
§ The electric field is perpendicular to the surface of both
spheres and points radially outward
20
Spherical Capacitor (3)
§ To calculate the electric field, we use a Gaussian surface
consisting of a concentric sphere of radius r such that r1 < r < r2
§ The electric field is always perpendicular to the Gaussian surface so
§ … which reduces to
…makes sense!
21
Spherical Capacitor (4)
§ To get the electric potential we follow a method similar to the one we
used for the cylindrical capacitor and integrate from the negatively
charged sphere to the positively charged sphere
§ Using the definition of capacitance we find
§ The capacitance of a spherical capacitor is
then
22
Capacitance of an Isolated Sphere
§ We obtain the capacitance of a single conducting sphere
by taking our result for a spherical capacitor and moving
the outer spherical conductor infinitely far away
§ Using our result for a spherical capacitor…
§ …with r2 = ∞ and r1 = R we find
…meaning V = q/4πε0R
(we already knew that!)
23