Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 2 for Electrical Engineering
Lecturers: Daniel Rohrlich , Ron Folman
Teaching Assistants: Ben Yellin, Yoav Etzioni
Week 5. Dielectrics, currents and resistance – Energy stored in a
capacitor • dielectrics • dielectrics and Gauss’s law • electrical
current • resistance and Ohm’s law • resistance and temperature •
electrical power
Sources: Halliday, Resnick and Krane, 5th Edition, Chaps. 29-31.
Energy stored in a capacitor
The flash on your camera uses a capacitor to store energy.
Although the flash is charged by a battery, the battery cannot
release the energy as quickly as the flash does.
How much energy can a capacitor store?
Energy stored in a capacitor
The flash on your camera uses a capacitor to store energy.
Although the flash is charged by a battery, the battery cannot
release the energy as quickly as the flash does.
How much energy can a capacitor store?
We have already learned that the potential energy U in a
configuration of electrical charges
does not depend on how the charges
got to where they are. So let’s
calculate U in a convenient way.
Energy stored in a capacitor
Step 1: The capacitor is uncharged. What is the potential
energy of the configuration of charges?
Energy stored in a capacitor
Step 1: The capacitor is uncharged. What is the potential
energy of the configuration of charges? U = 0
Energy stored in a capacitor
Step 1: The capacitor is uncharged. What is the potential
energy of the configuration of charges? U = 0
Step 2: The capacitor carries charges q' and –q' on its two
parts. How much work dW does it take to move an
infinitesimal amount of charge dq' from the negative part of
the capacitor to the positive part?
dq'
q'
–q'
Energy stored in a capacitor
Step 1: The capacitor is uncharged. What is the potential
energy of the configuration of charges? U = 0
Step 2: The capacitor carries charges q' and –q' on its two
parts. How much work dW does it take to move an
infinitesimal amount of charge dq' from the negative part of
the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq'
dq'
q'
–q'
Energy stored in a capacitor
Step 1: The capacitor is uncharged. What is the potential
energy of the configuration of charges? U = 0
Step 2: The capacitor carries charges q' and –q' on its two
parts. How much work dW does it take to move an
infinitesimal amount of charge dq' from the negative part of
the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq‘
Step 3: What is the total work done in charging the capacitor
to a charge q?
Energy stored in a capacitor
Step 1: The capacitor is uncharged. What is the potential
energy of the configuration of charges? U = 0
Step 2: The capacitor carries charges q' and –q' on its two
parts. How much work dW does it take to move an
infinitesimal amount of charge dq' from the negative part of
the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq‘
Step 3: What is the total work done in charging the capacitor
q
to a charge q? W = ∫ (q'/C) dq‘ = q2 /2C
Energy stored in a capacitor
Step 1: The capacitor is uncharged. What is the potential
energy of the configuration of charges? U = 0
Step 2: The capacitor carries charges q' and –q' on its two
parts. How much work dW does it take to move an
infinitesimal amount of charge dq' from the negative part of
the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq‘
Step 3: What is the total work done in charging the capacitor
q
to a charge q? W = ∫ (q'/C) dq‘ = q2 /2C
Step 4: What is the potential energy stored in the capacitor?
Energy stored in a capacitor
Step 1: The capacitor is uncharged. What is the potential
energy of the configuration of charges? U = 0
Step 2: The capacitor carries charges q' and –q' on its two
parts. How much work dW does it take to move an
infinitesimal amount of charge dq' from the negative part of
the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq‘
Step 3: What is the total work done in charging the capacitor
q
to a charge q? W = ∫ (q'/C) dq‘ = q2 /2C
Step 4: What is the potential energy stored in the capacitor?
U = q2 /2C = C(ΔV)2 /2
Energy stored in a capacitor
Example 1: You have three identical capacitors and a battery.
How should you combine them in one circuit so that the
capacitors will store the most energy?
Energy stored in a capacitor
Example 1: You have three identical capacitors and a battery.
How should you combine them in one circuit so that the
capacitors will store the most energy? Remember:
A
C
Ceff = 3C
Ceff = 3C/2
Ceff = 2C/3
B
D
Ceff = C/3
Energy stored in a capacitor
Example 2: A 3.55-μF capacitor C1 is charged to potential
difference of 6.30 V. It is then connected with a 8.95-μF
capacitor C2. (a) What is the final potential difference ΔV?
(b) What is the initial potential energy Uin? (c) What is the
final potential energy Ufin?
C1
C2
Energy stored in a capacitor
Example 2: A 3.55-μF capacitor C1 is charged to potential
difference of 6.30 V. It is then connected with a 8.95-μF
capacitor C2. (a) What is the final potential difference ΔV?
(b) What is the initial potential energy Uin? (c) What is the
final potential energy Ufin?
C1
C2
Answer: (a) Charge conservation!
Energy stored in a capacitor
Example 2: A 3.55-μF capacitor C1 is charged to potential
difference of 6.30 V. It is then connected with a 8.95-μF
capacitor C2. (a) What is the final potential difference ΔV?
(b) What is the initial potential energy Uin? (c) What is the
final potential energy Ufin?
C1
Answer: (a) Charge conservation!
q = (3.55 μF)(6.30 V)
= (C1 + C2) ΔV
q = 22.4 μC = (12.5 μF) ΔV
C2
ΔV = 1.79 V
Energy stored in a capacitor
Example 2: A 3.55-μF capacitor C1 is charged to potential
difference of 6.30 V. It is then connected with a 8.95-μF
capacitor C2. (a) What is the final potential difference ΔV?
(b) What is the initial potential energy Uin? (c) What is the
final potential energy Ufin?
C1
Answer: (b) Uin = C1(6.30 V)2/2
= 70.4 μJ
(c) Ufin = (C1 + C2) (ΔV )2/2
= 20.0 μJ
C2
Energy stored in a capacitor
Example 2: A 3.55-μF capacitor C1 is charged to potential
difference of 6.30 V. It is then connected with a 8.95-μF
capacitor C2. (a) What is the final potential difference ΔV?
(b) What is the initial potential energy Uin? (c) What is the
final potential energy Ufin?
C1
Answer: (b) Uin = C1(6.30 V)2/2
= 70.4 μJ
(c) Ufin = (C1 + C2) (ΔV )2/2
= 20.0 μJ
C2
A puzzle: Isn’t energy conserved?
Another puzzle:
A room with no windows contains three lamps. The switches to
the lamps are outside the room, next to the door – which is
closed. You can play with switches, turning the lamps on and
off, but you can enter the room just once. How can you find out
which switch controls which lamp, for all three lamps? (You
have no special equipment.)
Energy stored in a capacitor
Application: Where is the energy in a charged capacitor?
Consider a parallel-plate capacitor:
q2
q2
q 2d
U


.
2C 2 0 A / d  2 0 A
If we double the plate separation d, we double the energy. The
energy, therefore, seems to live between the plates, where the
electric field E = ΔV/d lives. If so, we can derive the energy
density uE of the electric field:
U
C (V ) 2  0 A / d ( Ed ) 2 1
uE 


 0E2
Ad
2 Ad
2 Ad
2
.
Dielectrics
A simple experiment with a dielectric (non-conducting)
material, e.g. glass:
C
q
C'
ΔV
ΔV '
voltmeter
q
Dielectrics
With the dielectric filling the capacitor (right), ΔV decreases to
ΔV ' = ΔV/κ where κ is called the dielectric constant of the
dielectric. Hence the capacitance increases from C to C ' = κC.
C
q
C'
ΔV
ΔV '
voltmeter
q
Dielectrics
The advantages of a dielectric in a capacitor include
1. increased capacitance,
2. increased maximum potential ΔV, and
3. mechanical support between plates (which helps increase the
capacitance even more since the plates can be very close).
Dielectrics
Phenomenological explanation:
Polar molecules (electric dipoles),
which orient themselves randomly
when E = 0, orient themselves in
the direction of the electric field
when E ≠ 0.
E=0
The net effect is an induced charge
inside the capacitor that decreases E
(and increases C)!
E≠0
Dielectrics
We can understand this explanation
better by considering what happens
when there is a conductor, instead
of a dielectric, between the plates of
a capacitor:
The conductor screens the electric
field via + and – charges on its
surface, thus reducing the potential
difference ΔV across the capacitor
and increasing the capacitance C.
E≠0
-
+
+
+
+
+
+
+
Dielectrics
Example 1: At right is a parallelplate capacitor of capacitance C. If
the conductor inserted between its
plates has the same cross-section
but only 1/3 the width, what is the
capacitance C ' of the capacitor
with the conductor inside?
E≠0
-
+
+
+
+
+
+
+
Dielectrics
Example 1: At right is a parallelplate capacitor of capacitance C. If
the conductor inserted between its
plates has the same cross-section
but only 1/3 the width, what is the
capacitance C ' of the capacitor
with the conductor inside?
Answer: The charge on the
capacitor is unchanged. The
potential difference is reduced by
2/3, i.e. ΔV ' = 2ΔV/3. Hence the
new capacitance is C ' = q/ΔV ' =
q/(2 ΔV/3) = 3(q/ΔV)/2 = 3C/2.
E≠0
-
+
+
+
+
+
+
+
Dielectrics
Example 2: At right is a parallelplate capacitor of capacitance C. If
the dielectric inserted between its
plates has the same cross-section
but only 1/3 the width, what is the
capacitance C ' of the capacitor
with the dielectric inside?
E≠0
-
+
-
+
-
+
Dielectrics
Example 2: At right is a parallelplate capacitor of capacitance C. If
the dielectric inserted between its
plates has the same cross-section
but only 1/3 the width, what is the
capacitance C ' of the capacitor
with the dielectric inside?
Answer: The charge on the
capacitor is unchanged. The
potential difference ΔV is reduced
to ΔV ' = (ΔV)(2/3 + 1/3κ). Hence
the new capacitance is C ' = q/ΔV '
= 3 κC/(2κ + 1).
E≠0
-
+
-
+
-
+
Dielectrics and Gauss’s law
Let’s apply Gauss’s law to the surface shown:
-
+
+
If there were no dielectric present, Gauss’s
law would yield ε0EA = q, where q is the total
+
charge on the inner surface of the capacitor
and E is the field inside. But with a dielectric
inside, the field inside is E ' and the total charge
inside the surface is q – q ', where –q ' is the charge on the

surface of the dielectric. Thus E '  dA 
E ' = E/κ, we have
   dA   E '  dA 
E
q  q'
0
q  q'
0
, but since

q
 0
.
Dielectrics and Gauss’s law
Although we obtained this version of Gauss’s law for the
specific case of a dielectric in a parallel-plate capacitor, it is the
correct generalization of Gauss’s law to include dielectrics:

E '  dA 
q
 0
.
Electrical current
We now go beyond electrostatics!
Definition: electrical current I is the flow of positive charge q
through a given cross-sectional surface. The magnitude of I is
dq
I
dt
and the unit of current is the ampere A = C/s.
This definition holds equally for electrons in a conductor and
protons in a beam. In a conductor, the electrons flow in the
direction opposite to the current and I = nAevdrift, where n is the
electron density in the conductor, A is the cross-sectional area
of the wire, –e is the charge of the electron, and vdrift is the
average speed of the electrons, which drift between collisions.
Electrical current
Example: The density of copper is 8.95 g/cm3, and home
copper wiring has a cross-sectional area A = 3.31 × 10–6 m2.
For a current in the wire of 10.0 A, what is the drift speed vdrift?
Note: each copper atom contributes one conduction electron.
Electrical current
Example: The density of copper is 8.95 g/cm3, and home
copper wiring has a cross-sectional area A = 3.31 × 10–6 m2.
For a current in the wire of 10.0 A, what is the drift speed vdrift?
Note: each copper atom contributes one conduction electron.
Answer: The atomic weight of copper (see the Periodic Table)
is 63.5 g/mol; using Avogadro’s number NA = 6.02214179 ×
1023 /mol, we find that for copper, the density of atoms (which
is also the density of conduction electrons) is n = (8.95 g/cm3) ×
(6.02 × 1023 /mol) / (63.5 g/mol) = 8.48 × 1022/cm3 electrons.
Electrical current
Answer: The atomic weight of copper (see the Periodic Table)
is 63.5 g/mol; using Avogadro’s number NA = 6.02214179 ×
1023 /mol, we find that for copper, the density of atoms (which
is also the density of conduction electrons) is n = (8.95 g/cm3) ×
(6.02 × 1023 /mol) / (63.5 g/mol) = 8.48 × 1022/cm3 electrons.
Now vdrift = I/neA and from our data we calculate
I
10.0 A

neA (8.48  10 28 /m3 )(1.602 10 –19 C )(3.31  10 – 6 m 2 )
 2.22  10 – 4 m/s .
So why doesn’t it take hours, after you flick the light switch, for
the light to go on?
Resistance and Ohm’s law
Ohm’s law is not fundamental; some materials obey it, some
don’t. Materials that obey Ohm’s law – including most metals
– are called ohmic. Ohm’s law, as we will study it, connects
two vector fields, the electric field E and the current density J,
via the conductivity σ of an ohmic material:
J = σE
.
(Here σ is not surface charge density.) If we integrate J across
any area element dA it crosses through, we get the current I,

J  dA  I ;
since charge does not build up in a conductor, I must be a
constant along the wire.
Resistance and Ohm’s law
Integrating E between two points on a wire, we obtain the
potential difference ΔV between those two points:

V  E  dr .
In the special case that E is constant along the wire and points
in the same direction as dr, the integral reduces to El, where l is
the length of the wire between the two points; then ΔV = El and
E = ΔV / l. From Ohm’s law we obtain I = σEA, where A is the
cross-sectional area of the wire, hence I = σ(ΔV)A / l. Now
defining the resistance of the wire to be R = l / σA, we finally
obtain
ΔV = IR ,
also called Ohm’s law. The unit of R is the ohm Ω = V/A.
Resistance and Ohm’s law
We also define the resistivity ρ of an ohmic material to be the
inverse ρ = 1/σ of the conductivity. So another expression for
the resistance R of the wire is R = lρ/A. The units of ρ are Ω·m.
Resistivities at 20°C.
Resistance and Ohm’s law
Example 1: Calculate the resistance per unit length of gold
wire having a radius of 0.321 mm, at 20°C.
Resistance and Ohm’s law
Example 1: Calculate the resistance per unit length of gold
wire having a radius of 0.321 mm, at 20°C.
Answer: Resistance per unit length is R/l = ρ/A. From the table
we find the resistivity of gold to be ρ = 2.44 × 10–8 Ω·m. The
cross-sectional area is A = π (0.321 mm)2 = 0.324 × 10–6 m2.
Therefore R/l = ρ/A = (2.44 × 10–8 Ω·m) / (0.324 × 10–6 m2) =
0.0753 Ω/m.
Resistance and Ohm’s law
Example 2: a current I flows through an ohmic wire that thins,
i.e. the cross-sectional area A gets smaller in the direction of I.
How do the electric field E, the current density J and the drift
velocity vdrift behave along the wire?
Resistance and Ohm’s law
Example 2: a current I flows through an ohmic wire that thins,
i.e. the cross-sectional area A gets smaller in the direction of I.
How do the electric field E, the current density J and the drift
velocity vdrift behave along the wire?
Answer: If a current I flows through the wire and charge does
not build up, then I is constant along the wire. But since A
decreases, J must increase. Since J = σE, also E must increase.
Finally, J is proportional to vdrift, so also vdrift must increase
along the wire.
Resistance and temperature
Over a limited range of temperature T, resistivity varies linearly
with T – T0 , where T0 = 20°C:
ρ = ρ0 [1 + α(T – T0)] ,
where ρ0 is the resistivity at 20°C and α is a constant that
characterizes the material.
Example 1: When does a light bulb carry more current – when
you just turn it on or when the glow is steady?
Resistance and temperature
Over a limited range of temperature T, resistivity varies linearly
with T – T0 , where T0 = 20°C:
ρ = ρ0 [1 + α(T – T0)] ,
where ρ0 is the resistivity at 20°C and α is a constant that
characterizes the material.
Example 2: The resistance of a platinum loop increases by
53.6% when it is dropped into melting indium. What is the
melting point of indium?
Resistance and temperature
Over a limited range of temperature T, resistivity varies linearly
with T – T0 , where T0 = 20°C:
ρ = ρ0 [1 + α(T – T0)] ,
where ρ0 is the resistivity at 20°C and α is a constant that
characterizes the material.
Example 2: The resistance of a platinum loop increases by
53.6% when it is dropped into melting indium. What is the
melting point of indium?
Answer: From 1.536 = R/R0 = ρ/ρ0 = 1 + α(T – 20°C) and the
value α = 3.92 × 10–3 /°C for platinum, we obtain the melting
temperature T = [0.536 / (3.92 × 10–3) + 20]°C = 157°C.
Electrical power
Remember the problem of charging one capacitor and then
connecting it to a second, uncharged capacitor? We found that
final potential energy was less than the initial potential energy.
Where did the energy go?
C1
C2
Electrical power
Remember the problem of charging one capacitor and then
connecting it to a second, uncharged capacitor? We found that
final potential energy was less than the initial potential energy.
Where did the energy go? To answer this question, let’s replace
one capacitor by a resistor.
R
At time t = 0 we close the switch. The
total potential around the circuit must
vanish for all t: q/C + IR = 0. Hence
dq/dt = I = –q/RC and we solve the
differential equation to obtain
q(t) = q(0) e–t/RC .
C
Electrical power
R
At time t = 0 we close the switch. The
total potential around the circuit must
vanish for all t: q/C + IR = 0. Hence
dq/dt = I = –q/RC and we solve the
differential equation to obtain
C
q(t) = q(0) e–t/RC .
The potential energy in the capacitor at time t is U = [q(t)]2/2C
= [q(0)]2 e–2t/RC/2C and the rate of loss of potential energy is
dU/dt = –[q(0)]2 e–2t/RC/RC2 = –I2R.
Conclusion: I2R is the rate of energy loss due to dissipation in
the resistor! It equals the power delivered to the resistor.
Halliday, Resnick and Krane, 5th Ed., Chap. 31, Exercise 28:
Part of an infinite array of identical 1-μΩ resistors is shown
below. A battery connects two remote junctions. Show that the
potential at any other junction is the average of the potentials at
the four nearest junctions.
Halliday, Resnick and Krane, 5th Ed., Chap. 31, Exercise 28:
Part of an infinite array of identical 1-μΩ resistors is shown
below. A battery connects two remote junctions. Show that the
potential at any other junction is the average of the potentials at
the four nearest junctions: V0 = (V1 + V2 + V3 + V4) / 4.
2
1
0
4
3
Halliday, Resnick and Krane, 5th Ed., Chap. 31, Exercise 28:
Answer: The total current into the junctions must vanish. Thus
I10 + I20 + I30 + I40 = 0. We also know that Vi = V0 + Ii0 R, where
R = 1 μΩ, for i = 1, 2, 3, 4. Summing V0 + Ii0 R over i we obtain
4V0 = V1 + V2 + V3 + V4 and so V0 = (V1 + V2 + V3 + V4) / 4.
2
1
0
4
3
from http://xkcd.com/356/
from http://xkcd.com/356/
from http://xkcd.com/356/