6/3/2013
Ch 30: Voltage and the Electric Field
DU = -Wc(i – f)
DV = DU
q
DV = -FDs
q
DV = -EqDs /q
[Work done BY the force]
DU = -FDs
F = Eq
DV = -EDs
E = electric field strength
Ds = change in distance
(negative of area under the E vs s curve)
Using the graph below, derive the formula of E(x),
V(x) and graph the V(x) function. (V(x) = -500x2)
• The electric field ALWAYS points along line of
decreasing voltage
• Often choose vf = 0 V (infinity for point
charges)
Find the electric potential (V) at any point “s”
inside a parallel plate capacitor. Let V=0 at the
negative plate. Also, remember that:
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Finding Electric Field from Potential
The voltage of a point charge is given below.
Use calculus to derive the formula for the
electric field for a point charge.
Assume electric field is constant
DV = -∫E ds
dV = -E ds
E = - dV
ds
(this is a derivative, a slope)
The on-axis potential of a ring of charge is
shown below. Use calculus to find the on-axis
electric field of a ring of charge.
Field lines
• Voltage (potential) is same along all points of a
field line
• Like contour lines on an elevation map
• E is perpendicular going “downhill” – towards
lower voltage
Given the following graph of the electric
potential of a region of space:
a. Calculate the electric field and draw a graph.
b. A proton is placed at 6 cm, will it move?
c. Are there any areas on the graph where the
proton will not move?
Given the following contour map of potential:
a. Draw lines to represent the direction of the
electric field at each point.
b. Indicate where the charge would be that is
producing this voltage (assume it is positive)
c. Estimate the strength of the field at each
point (use the distance between 100 V and 0
V)
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Electric Field and Shape
•
•
•
•
Electric field is zero inside a conductor
All charge resides on the surface
Exterior field is perpendicular to the surface
Charge can build up at point (metal in
microwave)
Capacitors (Condensers)
• Store electric charge for later use
– Camera flash
– Energy backup in computers
– Block surges of charge
– Stores “0”’s and “1”’s in RAM
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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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)
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)
Given the following spherical conductor, derive
the formula for the capacitance.
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Coaxial Cable
Determine the formula for the capacitance of a
coaxial cylindrical shell as shown. The formula
for electric field is: (substitute l = Q/L,
L=length of wire)
Capacitors in Parallel
Voltage is constant in parallel
Q = CV
Q = C1V + C2V
Q = CeqV
Ceq = C1 + C2 + C3 +…
C1
Capacitors in Series
Charge is constant in series
1= 1
Ceq C1
+
1
C2
+
1
C3
C1
C2
C3
+Q -Q
+Q -Q
+Q -Q
C2
(like using a larger plate)
Eq. Capacitance Ex. 1
Determine the equivalent capacitance for the
circuit below
if each capacitor is 5 mF.
C
Parallel first:
Ceq = C1 + C2
Ceq = 5mF + 5mF = 10 mF
1
C3
C2
Series:
1 = 1
Ceq 10
+
1
5
=
3/10
Ceq = 3.33 mF
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Given the following setup:
Given the circuit below:
a. Calculate the equivalent capacitance
b. Calculate the charge across each capacitor
c. Calculate the voltage on each capacitor
a. Calculate the charge on each capacitor (52.8
mC, 28.8 mC)
(HINT: Combine them for part a, and then split
them back up for b and c)
Parallel capacitors
Ceq = 5 mF +1 mF = 6 mF
Now both are series
1/ Ceq = 1/6 + 1/3
Ceq = 2 mF
Q = CV = (2 mF)(12 V) = 24 mC
For 3 mF capacitor
Q = CV
V = Q/C = 24 mC/3 mF = 8 V
For Combined Parallel Capacitors (same voltage)
V = Q/C = 24 mC/6 mF = 4 V
For 5 mF Capacitor
Q = CV = (5 mF)(4 V) = 20 mC
For 1 mF Capacitor
Q = CV = (1 mF)(4 V) = 4 mC
Given the following setup:
a. Calculate the equivalent capacitance. (2.0 mF)
b. Calculate the total charge that leaves the battery.
(8.0 mC)
c. Calculate the charge and voltage on each capacitor.
(Q1 = 8.0 mC, 2.7 V: Q2 = Q3 = 4.0 mC, 1.3 V)
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Given the following setup:
a. Calculate the equivalent capacitance. (0.609 mF)
b. Calculate the total charge that leaves the battery.
(14.6 mC)
c. Calculate the charge and voltage on each capacitor.
(Q1 = 14.6 mC, 14.6 V, Q2 = 14.6 mC, 7.30 V,
Q3 = 6.26 mC, 2.09 V, Q4 = 8.36 mC, 2.09 V )
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
Storage of Electric Energy
An electric device must hold 0.45 J of energy
while operating at 110 V. What size capacitor
should be chosen?
(ANS: 7.4 X 10-5 F, 74 mF)
Dielectrics
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
• Insulating paper or plastic
• Prevents charge from jumping the
gap
• Increases capacitance by a factor of
K
C = KCi
C = KeoA
s
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Dielectric Constants (K)
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
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
C = 2.0 X 10-11 F or 20 pF
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
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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.
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)
2. 5.0 kV
4. -550 V
6. 1.6 X 10-13 J
8. 1.5 V
10. 20 kV/m, 45o below the x-axis
12. -5000 V/m, 10,000 V/m, -5000 V/m
14. a) 0 V/m b) -200 V/m
18. 4.8 cm
20. 24 V
22. 32 mF
24. 50 mF
26.
28.
30.
32.
34.
36.
38.
40.
42.
44.
200 pF
1.41 kV
1/2
a) 1.11 X 10-7 J b) 0.71 J/m3
a) 20 kV/m
b) 20 pC
a) Higher at A
b) -70 V
a) Graphs
b) -2500x2 V
25 V
a) (0, -5, 10) V/m
b & c) graphs
a)
b)
46. a) 5000 V/m, 20,000 V/m b) graph
52.1.11 nC
58. 37 mF
60. 5.0 V, 15.0 V, 10.0 V
62. (45 mC, 9 V) (21.6 mC, 5.4 V) (21.6 mC, 3.6 V)
68.20 mF
70. 2.4 J
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