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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 = FDr cos0o = qEsi - qEsf
DU = qEsf – qEsi = 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.
(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 and
the final speed of the electron. (9.95 X 106 m/s)
Electric Potential: Voltage
• Voltage
• 1 Volt = 1 Joule/Coulomb
V = DU
q
Vab = Va – Vb = -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 or charge, the greater the
PE (DU = qV)
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)
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
U = qEs
V = DU/q
V = Es
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)
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. Calculate the charge on each plate (E = Q/e0A)
[6.95 X 10-10 C]
c. 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 is 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
• Energy an electron gains moving through a
potential difference of 1 V
1 eV = 1.6 X 10-19 J
Uelec = 1 q1q2
4peo r
k = 9.0 X 109 Nm2/C2
• Ex: An electron moving through 1000 V would
gain 1000 eV of energy
(derive from Force expression)
A proton is fired from “far away” at a 1.0 cm
diameter glass sphere of charge +100 nC.
a. Calculate the initial potential energy of the
system (just as it touches the sphere).
b. Since PE = KE, calculate the needed initial
speed of the proton.
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?
4.2 X 10-14 m (draw a comparision)
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An electron and a positron are created in the CERN
collider.
a. Calculate the potential energy they have when
they are 1.0 X 10-10 m apart.
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.
• Voltage is not directional (scalar)
• Charged particles (i.e.: electrons, protons) have a
voltage
Example 1
Point Charges: Example 2
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.
Voltage due to a Point Charge
V = kQ
r
Calculate voltage (electric potential) at point A as
shown below:
A
30 cm
52 cm
Q1 = +50 mC
Use Pythagoream theorem to calculate the distance
from A to Q2:
VA = V1 + V2
V1 = kQ = (9.00X 109 Nm2/C2)(5.00X10-5C)
r
(0.30 m)
V1 = 1.50 X 106 V
V1 = kQ = (9.00X 109 Nm2/C2)(-5.00X10-5C)
r
(0.60 m)
V2 = -7.5 X 105 V
A
30 cm
52 cm
Q1 = +50 mC
Q2 = -50 mC
Q2 = -50 mC
VA = 1.50 X 106 V -7.5 X 105 V
VA = 7.5 X 105 V
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Point Charges: Example 3
Use Pythagoream theorem to calculate the distance
from B to Q1 and to Q2:
Calculate voltage (electric potential) at point B as
shown below:
B
B
30 cm
26 cm
30 cm
26 cm
Q1 = +50 mC
Q2 = -50 mC
VA = V1 + V2
26 cm
Q1 = +50 mC
26 cm
Q2 = -50 mC
Point Charges: Example 4
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
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?
VA = 1.125 X 106 V –1.125 X 105 V
VA = 0 V
VQ = 3.6 X 105 V (This is the voltage caused by
the stationary charge)
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
VQ = kQ
r
=
(9.00X 109 Nm2/C2)(2.00X10-5C)
(0.500 m)
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?
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+
(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)
Anatomy of a capacitor
• Charge is stored on plates
• Charges does 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)
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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
A = area (larger, more storage)
d = distance
eo = 8.85 X 10-12 C2/Nm2
(permittivity of free space)
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)
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)
Dielectrics
• Insulating paper or plastic
• Prevents charge from jumping the
gap
• Increases capacitance by a factor of
K
C = KCi
C = KeoA
s
How Dielectrics Work
• Molecules orient themselves
and/or their electrons
• Decreases electric field
• Electric field and capacitance
are inversely related
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Capacitance and Studfinders
• Stud finder registers a change in capacitance.
• Wood acts as a dielectric
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.
C = KeoA
d
C = (3.7)(8.85 X 10-12 C2/Nm2 )(6.0X10-4
m2)
1.0 X 10-3 m
-11
C = 2.0 X 10 F or 20 pF
A 50- mF capacitor is charged to 160 V, It is then
disconnected from the battery and submerged in
water.
a. Calculate the charge stored on the plates
b. Calculate the new capacitance
c. Calculate the new voltage
d. Calculate the energy stored before and after
plunging it into the water.
Capacitance and Keyboards
• Pushing down decreases d
C = KeoA
d
• Change in capacitance detected
Calculate the charge that can be stored on the
capacitor at a voltage of 240 V.
Q = CV
Q = (2.0 X 10-11 F )(240 V)
Q = 4.8 X 10-9 C
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 battery. 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.
(7316 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. (9.06 X 10-9 F, 331
V, 5.0 X 10-4 J)
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Storage of Electric Energy
A camera flash stores energy in a 150 mF capacitor
at 200 V. How much energy can be stored?
U = ½ CV2
U = ½ (150 X 10-6 F)(200 V)2
U = 3.0 J
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.
b. Calculate the energy stored
c. Calculate the power is the capacitor is
discharged in 10 ms
(ANS: 7.4 X 10-5 F, 74 mF)
Charged Spheres
Act as point charges
At surface of sphere
A proton is released from rest at the surface of a
1.00 cm-diameter sphere charged to +1000 V.
a. Calculate the charge on the sphere (0.56 nC)
b. Calculate the speed of the proton when it is 1.00
cm from the sphere (note that it has potential
energy in both cases, U = qV) (3.1 X105 m/s)
Substitute:
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A thin ring has a radius R and charge Q. Find the
potential at a distance of z from the axis of the
ring.
A thin disk has a radius R and charge Q. Find the
potential at a distance of z from the axis of the
ring.
A 17.5 mm diameter dime is charged to +5.00 nC.
a. Calculate the potential of the dime (10,300 V)
b. Calculate the potential 1.00 cm above the dime
(3870 V)
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.
54. 4.0 X 107 m/s
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
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