Chapter 18: Electric potential and
! ! ! ! ! ! capacitance
What will we learn in this chapter?
Concept of electric potential and energy in electric interactions.
Electric work.
Voltage.
Equipotential surfaces in fields.
The Millikan oil-drop experiment.
Capacitors.
16.8F @ 350V
Circuit rules for capacitors.
Dielectrics.
Electric potential energy: classical analogy
Review of concepts learned in classical mechanics:
The work a particle moving from a to b along a straight line �s due
to a force F� does is given by
Wa→b = F s cos(φ)
where φ is the angle between the particle and the force.
Because the force field is conservative, we can express the work
done in terms of a potential:
Wa→b = Ua − Ub
When W is positive Ua > Ub and the potential energy
decreases (ball falls from high point). W = −∆Ugrav = mgh
Work-energy theorem: The change in kinetic energy
K is equal to the work done
Kb − Ka = Ub − Ua
Energy conservation:
Kb + Ub = Ka + Ua
Electric work
Example: Look at a charge in a uniform field.
The field generates a downward force.
The charge moves in turn a distance s from
a to b.
Work done by the field:
Wa→b = F s = qEs
The force is constant and independent of
the location.
In analogy to a gravitational field, if the field
and the path would be at an angle, we must
include the cosine of the angle.
y
Electric potential energy
Work:
Wa→b = F s = qEs
Definition in terms of a potential energy U:
y-component of the force: Fy = −qE
Since the field is uniform and only in the
y-direction, there are no other components.
Thus the force is independent of the taken
path.
Potential energy: U = qEy
When the charge moves from a to b we
obtain
Wa→b = Ua − Ub = qEya − qEyb = qE(ya − yb )
� ,U
If ya > yb the particle moves in the same direction as E
decreases, the field does positive work. Note, this is for q > 0.
Some conventional cases...
Some conventional cases...
Potential energy due to a point charge
Goal:
Compute the work done by a test charge q’ when it moves in the
electric field of a point charge q along the x axis from a to b.
The force changes
with the distance.
Perform the integral
�
W = F� d�s
Result:
W = kqq �
�
1 1
−
a
b
�
It follows for the potential
kqq �
kqq �
U
=
Ua =
b
b
a
Wa→b = Ua − Ub
Generalization
It can be shown that the work only depends on the distance between
points a and b and is independent of the taken path.
The potential energy U of a system consisting of a point charge q’
located in the field produced by a stationary point charge q at a
distance r from the charge is
qq �
U =k
r
Notes:
If a = b, then the work is zero.
This is characteristic of a conservative force field.
The potential energy is always defined w.r.t. some reference point
at which U = 0 (in general infinity).
U = 0 when q and q’ are infinitely apart, hence U is defined for q’
moving from an initial distance r to infinity.
Multiple point charges…
When we have more than one point charge causing the electric
� , the potential is simply the algebraic (not vector) sum of all
field E
contributions:
�
�
q1
q2
q3
q4
q5
�
U = kq
+
+
+
+
+ ...
r1
r2
r3
r4
r5
Any other equations regarding the
work W we derived still hold accordingly.
Because we can write any charge distribution
as a sum of individual point charges it follows
that we can always find the potential distribution
for any static electric field.
Every electric field due to a static charge
distribution is a conservative force field.
Potential
Analogous idea to the electric field (propensity of a charge to
generate a force on a test charge).
The potential energy per unit charge is called the ‘electric potential’.
The electric potential V at any point in an electric field is the electric
potential energy U per unit charge associated with a test charge q’ at
that point:
U
Wa→b
Ua
Ub
V = �
=
−
= Va − Vb
q
q�
q�
q�
It is a scalar quantity and the units are ‘Volts’ = Joule/Coulomb
1V = 1J/C
In the context of circuits the potential is often called voltage.
Example: a 9V battery has a potential difference (electric potential) of
9V between its two terminals.
Potential of a point charge
The potential V due to a point charge q at a distance r is given by the
potential energy U divided by the test charge q’:
V =
U
q
=
k
q�
r
Notes:
A similar expression holds for a collection
of point charges.
The unit of the electric field can also
be expressed as 1N/C = 1V/m.
Quiz:
Why is the bird not dead?
When is the bird dead?
Equipotential surfaces
Field lines help visualize electric fields…
Equipotential lines represent the potential at various points of an
electric field.
Definition: an equipotential surface is a surface
for which the potential is the same at every
point (think weather map).
single charge
dipole
two positive charges
Equipotential surfaces contd.
Note:
The potential energy for a test charge is the same at every point
on a given equipotential surface.
Because the potential does not change, the E-field
does not work on the test charge when it moves
on the equipotential surface.
The E-field can thus never have a component
tangent to the surface (otherwise the charge
would move).
The E-field must be perpendicular to the potential
surface at every point!
What about a uniform field?
Is a conducting surface equipotential?
Or…. Why is the electric field perpendicular to the surface of a
conductor?
We know the field is zero inside
the conductor (charges would
otherwise move).
Inside the tangent component
of the field is zero.
Due to energy conservation, the
tangential component of E just
outside the conductor must also
be zero. Otherwise a charge could
generate a net amount of work.
Hence E is perpendicular to a conducting
surface at every point.
A conducting surface is always an equipotential surface.
Electric field as a potential gradient
Concept:
(Equi)potential lines describe regions in the field where the
potential is the same.
Potential gradients describe the change in the potential from line
to line (think of acceleration vs velocity as field vs potential).
The magnitude of the E-field at any point of a equipotential surface
equals the rate of change of the potential ∆V with distance ∆s as the
point moves perpendicularly from the surface to an adjacent one a
distance ∆s away:
∆V
(potential gradient)
E=−
∆s
The negative sign tells us that when a point moves in the direction of
the E-field the potential decreases. Mathematically: Ei = ∂V /∂xi
� = ∇V
�
E
Millikan oil-drop experiment
How do we know charge is quantized?
How do we measure the charge of an individual electron?
Solution:
Millikan experiment performed 1909 - 1913 in Chicago.
Nobel prize in 1923.
First proof of subatomic
particles.
Millikan’s estimate of e:
e = 1.592 · 10−19 C
Todays estimate to 3 digits:
e = 1.602 · 10−19 C
Deviation: 0.62% !!!
Millikan oil-drop experiment contd.
0.0001 mm!
Known: potential Vab, density.
q = mg/E
E = Vab /d
m = (4/3)πr3 ρ
Unknown: radius of the drops
Millikan oil-drop experiment contd.
Compute the radius of the drop from the terminal speed and the
drag force (depends on r) as it falls when the field is turned off.
Repeat for thousands of drops…
Compute the charge.
Millikan always obtained integer multiples of a
given number
±e
± 2e
which thus was e.
± 5e . . .
Electronvolt (eV) – a common unit…
One can define an energy unit via the electron charge e.
Commonly used in calculations involving atomic, nuclear, and
condensed matter systems. Why? Because energies are small for
atomic particles.
The change in energy ∆U for an electron moving from a to b with a
potential difference ∆V = 1V is
∆U = e∆V = (1.602 · 10−19 C)(1V) = 1.602 · 10−19 J
Note:
In condensed matter physics 1eV/kB = 11604.5K .
13.6 eV are required to ionize hydrogen.
210 MeV are released in the fission of Pu-239.
Capacitors
Simplest definition:
A device that stores electric potential
energy and charge.
Applications:
Temporary battery (flash)
Pulsed power generators (tazer)
Signal processing
Sensing (fuel tanks in air planes)
Simple device:
Any two conductors separated
by a insulating medium.
Capacitors contd.
How do capacitors work?
Two separate conductors which are charged.
An electric field is generated between the conductors with a
potential difference.
The field between the conductors is proportional to the charge Q
on the conductors, i.e., the potential V is proportional to Q.
The capacitance C of a capacitor is the ratio of the magnitude of the
charge Q on either conductor to the magnitude of the potential
difference V between the conductors:
C = Q/V
The unit of capacitance is 1 Farad = 1 Coulomb/Volt: 1F = 1C/V
Circuit symbol:
Parallel-plate capacitors
Most common form:
Two parallel plates with area A,
separated by a distance d < A.
Uniform field between the
plates.
The E-field is proportional to
the charge per unit area of the plate
(surface charge density σ ):
σ = Q/A
E-field (via Gauss’s law):
E = σ/�0 =
Q
�0 A
Parallel-plate capacitors contd.
In general, the field between the plates is uniform.
The potential difference is given by
1 Qd
�0 A
Because C = Q/V it follows for a parallel-plate capacitor:
V = Ed =
The capacitance C of a parallel-plate capacitor in vacuum is
proportional to the area A of each plate and inversely proportional to
their separation d:
A
C = Q/V = �0
d
Note:
The capacitance depends only on geometric constants!
1F = 1C2 /J and �0 = 8.8542 · 10−12 F/m
Capacitors in series and in parallel
Why should we care?
Vendors make a finite set of capacitors, thus, if we want a given
capacitance, we need to combine different ones.
Two possible options:
Series connection (“one after the other”)
Parallel connection (“next to each other”)
Capacitors in series
Two capacitors are connected after each other (“series”) with V
constant.
Note that the charge on the upper plate of C2 and the lower plate
of C1 have to be of the same magnitude since c is not connected to
the power source.
Because of the conservation of the charge Q, the magnitude of the
charge on all plates must be the same in a series connection.
Capacitors in series contd.
Remember in general:
V =
Q is the same for all and so
Vac = V1 =
Q
C1
Q
C
Vcb = V2 =
Vab ≡ V = V1 + V2 =
Q
Q
+
C1
C2
Q
C2
We obtain for the equivalent capacitance of the circuit: Ceq =
1
1
1
=
+
Ceq
C1
C2
V = V1 + V2
Q
V
Q = Q1 = Q2
Summary: Capacitors in series
When capacitors are connected in series, the reciprocal of the
equivalent capacitance of a series combination equals the sum of the
reciprocal of the individual capacitances:
1
1
1
1
1
=
+
+
+
+ ...
Ceq
C1
C2
C3
C4
Furthermore
V = V1 + V2 + V3 + V4 + . . .
Q = Q1 = Q2 = Q3 = Q4 = . . .
The charges of the plates are all equal, but the potential on each
capacitor is different.
Capacitors in parallel
Parallel arrangement of capacitors with a constant potential V:
The charges on the plates are not necessarily equal and given by
Q1 = C1 V
Q2 = C2 V
The total charge is conserved and so Q = Q1 + Q2 = V (C1 + C2 ).
The equivalent capacitance is thus given by
Ceq = C1 + C2
Summary: Capacitors in parallel
When capacitors are connected in parallel, the equivalent capacitance
of a parallel combination equals the sum of the individual capacitances:
Ceq = C1 + C2 + C3 + C4 + . . .
Furthermore
V = V1 = V2 = V3 = V4 = . . .
Note:
In a parallel connection the equivalent capacitance is always
greater than any individual capacitance.
In a series connection, the equivalent capacitance is always less
than any individual capacitance.
Electric field energy
A common application of capacitors is to store energy.
When charged, capacitors act like stretched springs.
The work needed to transfer an additional small amount of charge
onto a capacitor is
q∆q
∆W = v∆q =
C
where v = q/C is the varying potential.
After an integration
�
q
1 Q2
U = Wtotal =
dq =
C
2 C
The energy in a capacitor is given by
1 Q2
1
U=
= CV 2
2 C
2
Energy density in an electric field
The energy stored in a capacitor is directly related to the electric
field between the capacitor plates.
One can say that the energy is stored in the field between the plates.
Calculation of the energy density u = U/Vol, Vol = Ad for a parallelplate capacitor:
1
2
U
2 CV
u=
=
Ad
Ad
Remember C = �0 A/d and V = Ed :
The energy density in a capacitor is given by
1
u = �0 E 2
2
This expression is valid for any field configuration in vacuum! Thus the
“vacuum” is not empty.
Dielectrics
Typical capacitor:
Two conducting plates
separated by a medium
and rolled up (space).
Role of the medium:
Keep plates separated at
a constant separation.
Prevent dielectric breakdown
(sparking) thus effectively
increasing the capacitance.
Dielectrics contd.
Dielectric constant K
The original capacitance is given by C0 = Q/V0.
When the dielectric is present C = Q/V.
Note that the charge does not change (why? conservation law!).
When the plates are completely filled by the dielectric, we define the
dielectric constant as
C
K=
C0
Note:
By definition K ≥ 1.
For vacuum K = 1 (good approximation for air).
Typical examples of K
For example, a capacitor with a plastic layer between the plates can
generally withstand about 3 times stronger potentials before
breaking down.
Electric fields in dielectrics
When a dielectric is inserted into a capacitor, the potential is
reduced by a factor K.
Because the electric field E is directly proportional to V, it must also
decrease by a factor K.
It follows:
E0
E=
K
where E0 is the field in vacuum.
Since E is smaller, the surface charge is smaller due to an induced
charge in the dielectric (also known as polarization, see Chapter 17).
Electric fields in dielectrics
original
induced
Molecular explanation of polarization
If the dielectric would be a conductor, charges would simply move to
the surfaces.
When we place molecules in a field, these have induced dipoles.
The redistribution of charge
causes the formation of a
net charge layer on the surface.
The internal field does not fully
cancel the field, because the charges cannot move freely.