Lecture 8
iClicker
§ A device has a charge q=10 nC and a potential V=100V,
what’s its capacitance?
§ A: 0.1 nF
§ B: 1nF
§ C: 10nF
§ D: 1010 F
§ E: 1F
iClicker
§ A device has a charge q=10 nC and a potential V=100V,
what’s its capacitance?
§ A: 0.1 nF
§ B: 1nF
§ C: 10nF
§ D: 1010 F
§ E: 1F
q = 10nC = 10
8
C
V = 100V
C = q/V = 10
8
/100 = 10
10
= 0.1nF
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.
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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
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
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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
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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.
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Spherical Capacitor (1)
§ Consider a spherical capacitor formed by two concentric
conducting spheres with radii r1 and r2
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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
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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!
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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
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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!)
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iClicker
q
§ A metal ball of radius R has a charge q.
§ Charge is changed q -> - 2q. How did it’s capacitance
change?
A: C->2 C0
B: C-> C0
C: C-> C0/2
D: C->4 C0
E: C->8 C0
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iClicker
q
§ A metal ball of radius R has a charge q.
§ Charge is changed q -> - 2q. How does it’s capacitance
changed?
A: C->2 C0
B: C-> C0
C: C-> C0/2
D: C->4 C0
E: C->8 C0
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Energy Stored in Capacitors
§ Capacitors store electric energy
1
U = qV
2
q = CV
1
2
U = CV
2
2
1q
U=
2C
(V created by “q”s, selfinteraction)
or
We want small
voltage, large
energy: large C
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Energy Density in Capacitors (1)
§ We define the energy density, u, as the electric potential energy
per unit volume
§ Taking the ideal case of a parallel plate capacitor that has no fringe
field, the volume between the plates is the area of each plate times
the distance between the plates, Ad
§ Inserting our formula for the capacitance of a parallel plate capacitor
we find
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Energy Density in Capacitors (2)
§ Recognizing that V/d is the magnitude of the electric
field, E, we obtain an expression for the electric potential
energy density for parallel plate capacitor
§ This result, which we derived for the parallel plate capacitor,
is in fact completely general.
§ This equation holds for all electric fields produced in any way
• The formula gives the quantity of electric field energy per unit
volume.
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Example: Isolated Conducting Sphere (1)
§ An isolated conducting sphere whose radius R is 6.85 cm has a
charge of q=1.25 nC.
Question 1:
How much potential energy is stored in the electric field of the charged conductor?
Answer:
Key Idea: An isolated sphere has a capacitance of C=4πε0R (see previous lecture).
The energy U stored in a capacitor depends on the charge and the capacitance
according to
… and substituting C=4πε0R gives
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Example: Isolated Conducting Sphere (2)
§ An isolated conducting sphere whose radius R
is 6.85 cm has a charge of q = 1.25 nC.
q
Question 2:
What is the field energy density at the surface of the sphere?
Answer:
Key Idea: The energy density u depends on the magnitude of the
electric field E according to
so we must first find the E field at the surface of the sphere. Recall:
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What is the total energy in E-field?
Z
1
2
dV = d⇤ sin d r dr
Utot =
udV =
2
= 4⇥r dr
R
Z 1
1
2 2
4⇡
✏0 E r dr =
R 2
◆2 2
Z 1✓
1
q 2
2⇡✏0
r
dr
=
4
4⇡✏
r
0
R
2
1 q
=
2 4⇡✏0 R
1
Yes!
qV
2
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Example: Thundercloud (1)
§ Suppose a thundercloud with horizontal dimensions of
2.0 km by 3.0 km hovers over a flat area, at an altitude of 500 m and
carries a charge of 160 C.
Question 1:
• What is the potential difference
between the cloud and the ground?
Question 2:
• Knowing that lightning strikes require
electric field strengths of approximately
2.5 MV/m, are these conditions sufficient
for a lightning strike?
Question 3:
• What is the total electrical energy contained in this cloud?
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Physics of a spark
V
E⇠
d1
+q
V
d
E⇠
V /d
V /d1
E0
e
E
-q
Ek ⇠ E ⇠ 1eV
⇠ 1µm
Espark ⇠ M eV /m
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Example: Thundercloud (2)
Question 1: What is the potential difference between the cloud and
the ground?
Answer:
§ We can approximate the cloud-ground system as a parallel plate
capacitor whose capacitance is
§ The charge carried by the cloud is 160 C
1q
V =
= 7.2 108
2C
…++++++++++++ …
++++++++++++
§ 720 million volts
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Example: Thundercloud (3)
Question 2: Knowing that lightning strikes require electric field
strengths of approximately 2.5 MV/m, are these conditions
sufficient for a lightning strike?
Answer:
§ We know the potential difference between the cloud and ground so
we can calculate the electric field
§ E is lower than 2.5 MV/m, so no lightning cloud to ground
• May have lightning to radio tower or tree….
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Example: Thundercloud (4)
Question 3: What is the total electrical energy contained in this cloud?
Answer:
§ The total energy stored in a parallel place capacitor is
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Electric circuits
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Circuit diagram
Lines represent conductors
The battery or power supply is represented by
The capacitor is represented by the symbol
Battery provides (a DC) potential difference V
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Charging/Discharging a Capacitor (2)
Illustrate the charging processing using a circuit diagram.
This circuit has a switch
• (pos c) When the switch is in position c, the circuit is open
(not connected).
• (pos a) When the switch is in position a, the battery is
connected across the capacitor. Fully charged, q = CV.
• (pos b) When the switch is in position b, the two plates of
the capacitor are connected. Electrons will move around the
circuit--a current will flow--and the capacitor will discharge.
c
c
8
demo
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I
-
V
+
I
+
V
31
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