Download electric potential energy

MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 12: ELECTRIC ENERGY
CLASSNOTES
ELECTRIC POTENTIAL
The unit of electric potential is the volt [V]. One joule [J] of work must be done by an external force
in order to move 1 coulomb [C] of charge from point A to a second point B:
[1 V = 1 J / C]
COULOMB’S CONSTANT (re-visited)
Coulomb’s constant k is not a fundamental constant of nature, but it is very useful. It is derived from
the vacuum permittivity constant o.
COULOMB’S CONSTANT (revisited)
k = (1/4) o
(o = 8.85 x 10-12 C2/N.m2)
This means “1 / 4o” can replace the constant “k” in electric equations.
Example 1. When moving an electrical charge from one point to another in the presence of an electrical
field, which quantity depends on the size of the charge that is moved?
(A) electric field (B) work done (C) potential difference (D) distance moved
(E) magnetic field
ELECTRIC POTENTIAL
The electric potential of a point charge, Q, is inversely proportional to the distance from the charge:
V = kQ / r
(k = 9.0 x 109 N.m2/C2)
Example 2. There is a hollow, conducting, uncharged sphere with a charge +Q inside the sphere.
Consider the potential and the electrical field at a point P1 inside the metal of the sphere.
Example 2. Diagram
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 2 (continued…)
At this point:
(A) only the electrical field is zero.
(B) only the electrical potential is zero.
(C) both the electrical field and the electrical potential are zero.
(D) neither the electrical field nor the electrical potential are zero.
(E) both the magnetic and electrical fields are zero.
2A.
(A) only the electrical field is zero.
Metal Sphere Electric Field
Metal Sphere Electric Field
[ E(r) = Electric field at radius, r
a = radius of metal sphere ]
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Metal Sphere Voltage
Metal Sphere Voltage
[ V(r) = Voltage at radius, r
a = radius of metal sphere ]
UNIFORM ELECTRIC FIELD
A charged particle inside a uniform electric field has potential energy UE due to its charge and potential
difference:
UE = qV
UNIFORM ELECTRIC FIELD
The electric potential (voltage) in a uniform electric field is directly proportional to distance. Uniform
electric fields are usually found inside parallel plate capacitors.
UNIFORM ELECTRIC FIELD
VOLTAGE EQUATION
V = Ed
( Voltage [V] = Electric Field [N/C] x Distance [m] )
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
PARALLEL PLATE CAPACITOR
PARALLEL PLATE CAPACITOR
The electric field between two parallel plates is uniform (the same) throughout the capacitor.
PARALLEL PLATE CAPACITOR
The capacitance C of a capacitor is the ratio of charge on either plate (conductor) to the potential
difference between the plates:
C=Q/V
[F] = [C] / [V]
PARALLEL PLATE CAPACITOR
The capacitance of an air-filled parallel plate capacitor is also determined by:
C = o A / d
(o = 8.85 x 10-12 C2/N.m2)
CAPACITOR POTENTIAL ENERGY
Potential energy is stored in the electric field inside a capacitor. It is measured in joules:
Uc = ½ CV2
Uc = Q2 / 2C
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Examples 3 - 5. A parallel plate capacitor is charged by a battery. The plate area is 10 cm x 200 cm,
the gap between plates is 0.50 mm, and the battery voltage is 20 V.
Examples 3 - 5. Diagram
3. Calculate the capacitance in nanofarads (1 nanofarad = 1 nF = 1 x 10-9 farads) assuming vacuum
between the plates.
3A.
(1) C = o(A/d)
(2) A = (L)(w)
(3) A = (0.1 m)(2.00 m)
(4) A = 0.2 m2
(5) C = (8.85 x 10-12)[(0.2 m2) / (5.0 x 10-4 m)]
(6) C = 3.54 x 10-9 F = 3.54 nF
4. Calculate the charge Q on the plates in nanocoulombs.
4A.
(1) C = Q / V
(2) Q = CV
(3) Q = (3.54 x 10-9 F)(20 V)
(4) Q = 7.08 x 10-8 C = 70.8 nC
5. Evaluate the stored energy in nanojoules.
5A.
(1) Uc = ½ CV2
(2) Uc = ½ (3.54 x 10-9 F)(20 V)2
(3) Uc = 7.08 x 10-7 J = 708 nJ
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CAPACITORS IN SERIES
CAPACITORS IN SERIES
The equivalent capacitance of a series combination is computed:
1/Cs = 1/C1 + 1/C2 …
1 / Cs = (1 / Ci)
Examples 6 - 9. Two capacitors are connected in series.
Examples 6 – 9. Diagram
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6. What is the equivalent capacitance in microfarads?
6A.
(1) 1 / Cs = 1 / C1 + 1 / C2
(2) 1 / Cs = 1 / (3 F) + 1 / (6 F)
(3) 1 / Cs = 2 / (6 F) + 1 / (6 F)
(4) 1 / Cs = 3 / (6 F)
(5) 1 / Cs = 1 / (2 F)
(6) Cs = 2 F
7. Determine the charge on the 3 microfarad capacitor in microcoulombs.
7A.
(1) Cs = Q1 / V
(2) Q1 = (Cs)(V)
(3) Q1 = (2 F)(10 V)
(4) Q1 = 20 C
8. Calculate the voltage across the 3 microfarad capacitor.
8A.
(1) C1 = Q1 / V1
(2) V1C1 = Q1
(3) V1 = Q1 / C1
(4) V1 = 20 C / 3 F
(5) V1 = 6.67 V
9. Calculate the voltage across the 6 microfarad capacitor.
9A.
(1) C2 = Q2 / V2
(2) V2C2 = Q2
(3) V2 = Q2 / C2
(4) Q2 = (Cs)V (V = battery voltage)
(5) Q2 = (2 F)(10 V)
(6) Q2 = 20 C
(7) V2 = 20 C / 6 F
(8) V2 = 3.33 V
VOLTAGE IN SERIES CAPACITORS
The potential difference (voltage) across a series of capacitors is computed:
V = V1 + V2 …
CHARGE IN SERIES CAPACITORS
Capacitors connected in series have the same charge Q.
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VAN NUYS HIGH SCHOOL
CAPACITORS IN PARALLEL
PARALLEL CAPACITORS
The equivalent capacitance of a parallel combination is determined:
Cp = C1 + C2 …
Cp = Ci
VOLTAGE IN PARALLEL CAPACITORS
The potential difference (voltage) across all capacitors in a parallel group remains constant.
Example 10. What is the equivalent capacitance of the combination shown?
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10A.
(1) Left side: CL = 10 F + 20 F
(2) CL = 30 F
(3) Right side: CR = 30 F + 30 F
(4) CR = 60 F
(5) Both sides: 1 / Ceq = (1 / CL) + (1 / CR)
(6) 1 / Ceq = 1/30 F + 1/60 F
(7) 1 / Ceq = 2/60 F + 1/60 F
(8) 1 / Ceq = 3/60 F = 1/20 F
(9) Ceq = 20 F
CHARGE IN PARALLEL CAPACITORS
The total charge stored by a group of capacitors connected in parallel is computed:
Q = Q1 + Q2 …
Examples 11 - 14. Two capacitors are connected in parallel.
11. What is the voltage and charge for the 2 microfarad capacitor?
11A.
(1) Since the capacitors are connected in parallel, voltage remains constant at 10 V
(2) C = Q / V
(3) Q = CV
(4) Q = (2 F)(10 V)
(5) Q = 20 C
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12. What is the voltage and charge for the 4 microfarad capacitor?
12A.
(1) Voltage is constant at 10 V
(2) C = Q / V
(3) Q = CV
(4) Q = (4 F)(10 V)
(5) Q = 40 C
13. What is the total charge stored?
13A.
(1) Q = Q1 + Q2
(2) Q = 20 C + 40 C
(3) Q = 60 C
14. Determine the equivalent capacitance of the combination in microfarads.
14A.
(1) Ceq = C1 + C2
(2) Ceq = 2 F + 4 F
(3) Ceq = 6 F
DIELECTRIC MATERIALS
When a dielectric (insulating material) is inserted between the plates of a capacitor,
the capacitance of the device increases by a factor .
DIELECTRIC MATERIALS EQUATION
C = o(A / d)
(the value  depends on the insulator)
Example 15. A 200 V battery is connected to a 0.50 F, parallel plate, air-filled capacitor. Now the
battery is disconnected, with care taken not to discharge the plates. Some Pyrex glass is next inserted
between the two plates, completely filling up the space. What is the final potential difference between
the plates? (dielectric constant  = 5.60)
15A.
(1) Determine initial charge: C = Q / V
(2) Q = CV
(3) Q = (5.0 x 10-7 F)(200 V)
(4) Q = 1.00 x 10-4 C
(5) Determine new capacitance: C’ = o(A/d) = C
(6) C’ = (5.60)(5.0 x 10-7 F)
(7) C’ = 2.80 x 10-6 F
(8) Determine new voltage: C’ = Q / V’
(9) V’ = Q / C’
(10) V’ = (1.00 x 10-4 C) / (2.80 x 10-6 F)
(11) V’ = 35.7 V
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 16. A parallel plate capacitor has dimensions 2.00 cm x 3.00 cm. The plates are separated by
a 1.00 mm thickness of paper (dielectric constant  = 3.70). What is the charge that can be stored on
this capacitor, when connected to a 9.00 volt battery?
16A.
(1) C = o(A / d)
(2) A = (2.00 x 10-2 m)(3.00 x 10-2 m)
(3) A = 6.00 x 10-4 m2
(4) C = (3.70)(8.85 x 10-12)(6.00 x 10-4 m2 / 1.00 x 10-3 m)
(5) C = 1.97 x 10-11 F
(6) C = Q / V
(7) Q = CV
(8) Q = (1.97 x 10-11 F)(9.00 V)
(9) Q = 1.77 x 10-10 C = 17.7 nC
Example 17. Very large capacitors have been considered as a means for storing electrical energy. We
constructed a very large parallel plate capacitor of plate area 1.00 m2 using pyrex ( = 5.60) of thickness
2.00 mm as a dielectric. How much electrical energy would it store at a plate voltage of 6000 V?
17A.
(1) Determine the capacitance: C = o(A / d)
(2) C = (5.60) (8.85 x 10-12)(1.0 m2 / 2.00 x 10-3 m)
(3) C = 2.48 x 10-7 F
(4) Determine electrical energy: U = ½ CV2
(5) Uc = ½ (2.48 x 10-7 F)(6000 V)2
(6) Uc = 0.45 J
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