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Electric Potential Energy
Voltage and Capacitance
Chapter 17
Potential Energy of a charge
• Wants to move when it has
high PE
• Point b
– U = max
– K = min
• Point a
– U = min
– K = max
Electric Work
Charge moving between plates.
Work = FDs cos0o = qEsi - qEsf
DU = qEsi – qEsf = qEDs
W = -DU
A 2.0 cm diameter disk capacitor has a 2.5 mm
spacing and an electric field of 2.70 X 105 N/C.
a. An electron is released from rest at the negative
plate. Calculate the change in potential energy
(DU). (-1.08 X 10-16 J)
b. Calculate the final speed of the electron. (1.54
X 107 m/s)
The electric field of a capacitor is 2.82 X 10 5 N/C.
The spacing between the plates is 2.00 mm.
a. A proton is released from rest at the positive plate.
Calculate the change in potential energy. (-9.02 X
10-17 J)
b. Calculate the final speed of the proton. (3.29 X 105
m/s)
c. An electron is released from the halfway point
between the plates. Calculate the change in PE
(DU) and the final speed of the electron. (9.95 X 106
m/s, left)
Electric Potential: Voltage
• Voltage
• 1 Volt = 1 Joule/Coulomb
DV = DU
q
DV = -Wba
q
Work done by the
electric field to
accelerate the
charge
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• The higher the rock, the greater the PE
• The greater the Voltage difference, the greater the
PE (DU = qV)
• Great Q, greater DU
An electron is accelerated in a TV tube
through a potential difference of
5000 V.
a. Calculate the change in PE of the
electron (-8.0 X 10-16 J)
b. Calculate the final speed of the
electron (m = 9.1 X 10-31 kg) (4.2
X 107 m/s (1/7th speed of light)
c. Calculate the final speed of a
proton (mass =
1.67 X 10-27 kg) (9.8 X 105 m/s
(0.3% speed of light)
Equipotential Lines
• Equipotential lines are
perpendicular to electric field
lines
• Voltage is the same along
equipotential lines
• Like contour (elevation) lines on
a map
Electric Field and Voltage
Equipotential lines for point charges
Voltage increases as you move between plates
DU = qEDs
DV = DU/q
DV = EDs
Greater the distance between plates, the greater the
voltage
The greater the E field, the greater the voltage
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Calculate the electric field between two plates
separated by 5.0 cm with a voltage of 50V. (1000
V/m)
An electron in a television set is accelerated
through a 2.86 X 104 V/m electric field. The
screen is 35 mm from the cathode.
a. Calculate the net change in the potential energy
of the electron during the acceleration process (1.6 X 10-16 J)
b. How much work is done by the electric field in
accelerating the electron? (1.6 X 10-16 J)
c. What is the speed of the electron when it strikes
the screen? (1.87 X 107 m/s)
A capacitor is constructed of 2.0 cm diameter
disks separated by a 2.0 mm gap, and charged
to 500 V.
a. Calculate the electric field strength (V = Es)
[2.5 X 105 N/C]
b. A proton is shot through a hole in the negative
plate towards the positive plate. It has an initial
speed of 2.0 X 105 m/s. Does it have enough
energy to reach the other side? (V = DU/q) [DU
= 8 X 10-17 J, K = 3.34 X 10-17 J]
Electron Volt
Potential Energy of point charges
Uelec = k q1q2
r
k = 9.0 X 109 Nm2/C2
• Energy an electron gains moving through a
potential difference of 1 V
1 eV = 1.6 X 10-19 J
• Ex: An electron moving through 1000 V would
gain 1000 eV of energy
A proton is fired from “far away” at a 1.0 cm
diameter glass sphere of charge +100 nC.
a. Calculate the potential energy of the system (just
as it touches the sphere). (2.88 X 10-14J)
b. Since PE = KE, calculate the needed initial
speed of the proton. (5.87 X 106 m/s)
(derive from Force expression)
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In Rutherford’s gold foil experiment, he fired alpha
particles (+2 charge, 6.64 X 10-27 kg) at 1.61 X
107 m/s at a gold nucleus (+79 charge). How
close could the alpha particle get to the nucleus?
An electron and a positron are created in the CERN
collider.
a. Calculate the potential energy they have when
they are 1.00 X 10-10 m apart. (2.30 X 10-18J)
b. Calculate the velocity they need to escape from
one another. Remember that PE = KE, but you
will need to consider the KE of both particles
added together. (1.59 X 106 m/s)
4.2 X 10-14 m
Voltage due to a Point Charge
Example 1
• Voltage is not directional (scalar)
• Charged particles (i.e.: electrons, protons) have a
voltage
Consider a +1.0 nC charge.
a. Calculate the electric potential (voltage) at a
point 1.0 cm from the charge
b. Calculate the electric potential at a point 3.0 cm
from the charge.
V = kQ
r
Point Charges: Example 2
Use Pythagorean theorem to calculate the distance
from A to Q2:
Calculate voltage (electric potential) at point A as
shown below:
A
A
30 cm
30 cm
52 cm
Q1 = +50 mC
52 cm
Q2 = -50 mC
Q1 = +50 mC
Q2 = -50 mC
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VA = V1 + V2
V1 = kQ = (9.00X 109 Nm2/C2)(5.00X10-5C)
r
(0.30 m)
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V1 = 1.50 X 10 V
V1 = kQ = (9.00X 109 Nm2/C2)(-5.00X10-5C)
r
(0.60 m)
V2 = -7.5 X 105 V
Suppose two helium nuclei (+2 each) are brought to
within 5.00 X 10-14 m of each other.
a. Calculate the potential energy of the system. (1.8 X 1014 J)
b. Calculate the work needed to bring them together from
very far away (infinity). (1.8 X 10-14 J)
c. Calculate the speed of the nuclei if they are released
and allowed to move far away from each other. The
mass of one nuclei is 6.64 X 10-27 kg. (1.65 X 106 m/s)
d. Calculate the voltage produced by just one nuclei at the
distance of 5.00 X 10-14 m. (57.6 kV)
VA = 1.50 X 106 V -7.5 X 105 V
VA = 7.5 X 105 V
A positive +5.00 mC charge is placed at the origin. A -10.00
mC charge is placed at (0.200, 0.000)m.
a. Calculate the voltage (potential) at (0.100, 0.000) m.
[-4.5 X 105 V]
b. Calculate the point in between the particles where the
voltage is zero. [0.067 m, 6.67 cm]
c. Calculate the magnitude of the potential energy between
the two particles.
[2.25 J]
d. Suppose the -10.00 mC charge is fixed in place. Calculate
the speed of +5.00 mC (m = 0.500 g) particle when it
collides with it. [94.9 m/s]
Point Charges: Example 3
Calculate voltage (electric potential) at point B as
shown below:
B
30 cm
26 cm
26 cm
Q1 = +50 mC
Use Pythagoream theorem to calculate the distance
from B to Q1 and to Q2:
30 cm
Q1 = +50 mC
VA = V1 + V2
V1 = kQ = (9.00X 109 Nm2/C2)(5.00X10-5C)
r
(0.40 m)
V1 = 1.125 X 106 V
V1 = kQ = (9.00X 109 Nm2/C2)(-5.00X10-5C)
r
(0.40 m)
V2 = -1.125 X 105 V
B
26 cm
Q2 = -50 mC
26 cm
Q2 = -50 mC
VA = 1.125 X 106 V –1.125 X 105 V
VA = 0 V
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Calculate voltage (electric potential) at point B as
shown below (-188 kV or -1.88 X 105 V)
How much work is required to bring a charge of q =
3.00 mC to a point 0.500 m from a charge Q =
20.0 mC?
B
VQ = kQ
r
30 cm
30 cm
20 cm
Q1 = +50 mC
Q2 = -50 mC
W = DU (like the work to lift a book to a shelf)
V = DU
q
V=W
q
W = Vq
W = (3.6 X 105 V )(3 X 10-6 C)
W = 1.08 J
+
Point Charges: Example 4
(9.00X 109 Nm2/C2)(2.00X10-5C)
(0.500 m)
VQ = 3.6 X 105 V (This is the voltage caused by
the stationary charge)
Point Charges: Example 5
Which of three sets of charges has the most:
• positive potential energy?
• The most negative potential energy?
• Would require the most work to separate?
(i)
-
=
Capacitors (Condensers)
• Store electric charge for later use
+
-
+
+
(ii)
–
–
–
–
Camera flash
Energy backup in computers
Block surges of charge
Stores “0”’s and “1”’s in RAM
(iii)
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Anatomy of a capacitor
• Charge is stored on plates
• Charges do not jump the gap
• Often rolled to increase surface area
Capacitance
Q = DVC
Q = charge stored on the plate
DV = Voltage
C = Capacitance [Farad (F)]
Most capacitors between 10-12 F and 10-6 F
(picoFarad to microFarad)
C = eoA
d
A = area (larger, more storage)
d = distance
eo = 8.85 X 10-12 C2/Nm2
(permittivity of free space)
Calculate the capacitance of a capacitor whose
plates are 20 cm X 3.0 cm and are separated by
a 1.0 mm air gap.
a. Calculate the capacitance using: (53 pF)
C = eoA
d
b. If the capacitor is attached to a 12-Volt battery,
what is the charge in each plate? (6.4 X 10-10 C)
c) Calculate the electric field between the plates.
(1.2 X 104V/m)
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The spacing between the plates of a 1.00 mF
capacitor is 0.050 mm.
a. Calculate the surface area of the plates (5.65
m2)
b. Calculate the charge if the plate is attached to
a 1.5 V battery. (1.5 mC)
Dielectric Constants (K)
Dielectrics
• Insulating paper or plastic
• Prevents charge from jumping the
gap
• Increases capacitance by a factor of
K
C = KCi
C = KeoA
d
How Dielectrics Work
• Molecules orient themselves
and/or their electrons
• Decreases electric field
• Electric field and capacitance
are inversely related
Capacitance and Studfinders
• Stud finder registers a change in capacitance.
• Wood acts as a dielectric
Capacitance and Keyboards
• Pushing down decreases d
C = KeoA
d
• Change in capacitance detected
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A parallel plate capacitor has plates 2.0 cm by
3.0 cm. The plates are separated by 1.0 mm
thickness of paper (K=3.7). Calculate the
capacitance.
Calculate the charge that can be stored on the
capacitor at a voltage of 240 V.
Q = 4.8 X 10-9 C or 4.8 nC
C = 2.0 X 10-11 F or 20 pF
A 50.0 mF capacitor is charged to 160.0 V, It is then
disconnected from the battery and submerged in
water.
a. Calculate the charge stored on the plates [8.00 C]
b. Calculate the new capacitance [4.00 F]
c. Calculate the new voltage [2.00 V]
d. Calculate the energy stored before and after
plunging it into the water. [640J, 8.00 J]
Storage of Electric Energy
A parallel plate capacitor is filled with a
dielectric of K = 3.4. The plates are square and
have a side length of 2.0 m. It is connected to a
100 V source. The plates are separated by
4.02 mm.
a. Calculate the capacitance (3.0 X 10-8 F)
b. Calculate the charge on the plates (3.0 X 10-6 C)
c. Calculate the electric field between the plates.
(2.5 X 104 V/m)
d. Calculate the energy stored (1.50 X 10-4 J)
e. The battery is disconnected and the dielectric
removed. Calculate the new capacitance,
voltage and energy stored. (8.8 X 10-9 F, 341 V,
5.1 X 10-4 J)
A camera flash stores energy in a 150 mF capacitor
at 200 V. How much energy can be stored?
U = 3.0 J
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An electric device must hold 0.45 J of energy
while operating at 110 V. What size capacitor
should be chosen?
A 2.0 mF capacitor is charged to 5000 V
a. Calculate the charge on one of the plates. [0.010C]
b. Calculate the energy stored [25 J]
c. Calculate the power if the capacitor is discharged
in 10 ms [2.5 MW]
(ANS: 7.4 X 10-5 F, 74 mF)
2. 2.7 X 106 m/s
4. 25,000 m/s
6. -9 X 10-7J
10. 1.874 X 107 m/s
12. -8.4 X 104 V
14. -0.712 V
16. a) 1833
b) 1
18. a) 1.5 V
b) 8.3 X 10-12 C
20. a) 200 V
b) 400 V
22. 0.23 m, 0.056 m
24. -5.8 kV
26.
28.
30.
32.
34.
36.
40.
42.
44.
46.
1410 V
3140 V, 10.0 nC
+ 12 cm
a) Both positive b) (draw graph)
-10 nC, +40 nC
a) zero at +∞
b) (same as (a))
0.72 J
b) 14.4 N c) 21.9 m/s & 11.0 m/s
1.01 X 105 m/s
b) 9.6 X 10-16x2 J
b) 1.92 X 10-15 N/m
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a) 2.1 X 10 V/m b) 9.4 X 107 m/s
54. 4.0 X 107 m/s
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One Charge
Two Charges
Python Challenge:
#We will model a charged particle released from one
plate of a capacitor. We will assume it races toward
the other plate.
Allow the user to input:
Voltage between plates
Charge and mass of the particle
Output:
Speed when it hits the other plate
Can you modify the program to run multiple times?
Can you build error checking into the program?
W = -DU
DV = DU
q
DV = EDs
Uelec = k q1q2
r
V = kQ
r
Q = DVC
me = 9.11 X 10-31 kg
k = 9.0 X 109 Nm2/C2
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