Download Kelly Nguyen Electrical Energy And Capacitors Definitions

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Electrical Energy And Capacitors

Definitions:
o Electrical Potential Energy: Work that could be done by the electric force (units
= Joules) W = ∫FE ∙ dx
o Electrical Potential: The amount of PE per unit charge. V = U/q
o Voltage: The difference between two electrical potentials. VAB= VA- VB
o Potential Difference: Same as Voltage

Coulomb’s Law: Felec = kQ1Q2/r2

F = Eq = kQ/r2 ; V = Ed ; V = ∫E ∙ dl
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Electric field lines is defined as the way positive moves
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Flux is the amount of field lines penetrating an area.
o Gauss’s Law: ∫E ∙ dA = Qenc/ε0

Point Charge:
∫E ∙ dA = Qenc/ε0
E 4πr2 = Qenc/ε0
E = Qenc/ε04πr2
Red circle = Gaussian Surface
+Q

Charged cylinder: λ= Q/l
∫E ∙ dA = Qenc/ε0
E(2πr2 + 2πrl) = Qenc/ε0 Note: 2πr2 = 0 because no field lines at end
caps
E 2πrl = λl/ε0
E = λ/2πr ε0
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Red Cylinder = Gaussian Surface

Sheet of charge: σ = Q/A
∫E ∙ dA = Qenc/ε0
E 2A = σA/ ε0
E = σ/2 ε0
Prism = Gaussian Surface
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+
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+
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Non conducting sphere: Total charge = uniform charge Q
o

+
∫E ∙ dA = Qenc/ε0=
ρV/ ε0 where ρ = Q/V
Capacitance is defined as Q/V+AB
+
+
+
o To derive any of the equations,
use Gauss’s Law and V = ∫E ∙ dl

+
Parallel Plate capacitor:
+
∫E ∙ dA = Qenc/ε0
E A = σA/ ε0
E = σ/ε0
V = ∫E ∙ dl = ∫ σ/ε0 dl = σd/ε0
C= Q/V = Q/ σd/ε0 = ε0A/d
+Q
d
-Q
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
Spherical Capacitor:
∫E ∙ dA = Qenc/ε0
E 4πr2 = Qenc/ε0
E = Qenc/ε04πr
RA
V = ∫E ∙ dr
RA RB
RA
2
RB
V= ∫ Qenc/ε04πr2 ∙ dr
RB
RA
V=Qenc/ε04πr2 ∙ r
RA RB
V=Qenc/ε04πr
RB
V= Qenc/ε04π (-1/RB + 1/RA)
V = Qenc/ε04π(1/RA - 1/RB)
C= Q/ Qenc/ε04π(1/RA - 1/RB)
C= ε04π/(1/RA - 1/RB)

Cylindrical Capacitor:
∫E ∙ dA = Qenc/ε0
-
-
E 2πrl = λl/ε0
-
-
E = λ/2πr ε0
RA
V = ∫E ∙ dr
RA RB
-
-
E(2πr2 + 2πrl) = Qenc/ε0
V= ∫ λ/2πrε0 ∙ dr
RB
RA
V= λ/2πε0 ∫ 1/r ∙ dr
RB RA
V= (λ/2πε0)ln r
RB
+
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V = (λ/2πε0) ln (RB – RA)
V = (λ/2πε0) ln(RB/RA)
C = Q/(λ/2πε0) ln(RB/RA)
C = Q/(Q/2πlε0) ln(RB/RA)
C = 2πlε0/ln(RB/RA)

Capacitors in series:
o Charges: Same- Q = Q1 = Q2 = QT
o All voltages add up: VT = V1 + V2
o Capacitance: the inverse of the total is equivalent to the sum of the inverses of
the other capacitors 1/CT = 1/C1 + 1/C2…1/C(N-1) + 1/CN

Capacitors in Parallel:
o Charges: Add up QT = Q1 + Q2
o Voltages: Same VT = V1 = V2
o Capacitors: Add up CT = C1 + C2 … C(N-1) + CN

Dielectric: An insulator in order to prevent electricity through an area
W= QV
Q
Q
W= ∫V∙ dq = W = ∫Q/C∙ dq = Q2/2C
0
0
U = ½ Q2/C= ½ Q2/Q/V = ½ VC x V = ½ V2C where Q = CV
U = ½ V2C =( ½ ε0A/d V2)/Ad = ½ ε0E2
C= k ε0A/d
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Sample AP MC Questions:
1). A capacitor is dependent on the charge and voltage. As the charge increases or
voltage decreases, the capacitance increases. If a charge decreases or voltage
increases, then the capacitance increases. A parallel plate capacitor is inversely
proportional to which of the following:
A). Distance
B). The smaller radii
C). The larger radii
D). Both B and C
E). Area
2). An electric field exists in the region of space shown above. The angle between the
electric field and voltage is 60º where the voltage = 500 V. If the distance is 5m,
what is the magnitude of the electric field?
A). 250 V/m
B). 50 V/m
C). 200 V/m
D). None of the above
3). If the force of electricity = 5 N, voltage is equivalent to 10 V, and d = 2m, what is
the charge?
A). 1 C
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B). 3 C
C). 5 C
D). 2 C
4). If the voltage is defined as V = 10x2 + 3y + 2z2, which of the following is true?
A). The z component of this field is constant .
B). The y component of this field is constant.
C). A proton within this field would feel a positive electric field in the x direction.
D). A proton within this field would feel a positive electric field in the z direction.
5). In the picture below, +q and +2q is separated by the distance D, where at point D,
V= 0 and E = 0.
D
+Q
+2Q
What is the magnitude of the radius?
A). ½ r
B). 1/3 R
C). ¼ R
D). 0
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Sample Open Response:
Consider the circuit below:
C1
C2
C3
C4
C5
C6
V
a). What is the total capacitance of C1, C2, C3, C4? What is the total capacitance of the
whole circuit? Express answer is terms of variables given.
b). What would be the energy density of the whole circuit?
c). What is the total charge?
d). The voltage present is 100 V.A dielectric with a constant k = 1.0 x 107 is present in
C6. If the distance = 5m, what is the Area of C6?
e). Determine the total force of electricity if the Electric field = 5000 V /m
Investigation:
During the electrostatics lab, observations were noted when a charged piece of plastic
and rod were near a neutral pith ball. When the charged piece of plastic was near the neutral pith
ball, the ball is attracted to the plastic; however, when the oppositely charged piece of plastic is
near the neutral pith ball, then the ball repels. This is due to the charges within the neutral pith
ball. The charged piece of plastic will attract the opposite charges on the pith ball. Since the
charges differ, they will attract. Once the oppositely charged piece of plastic goes near the
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already charged pith ball, they will repel. When the charged rod was used, the neutral ball goes
back to the rod. When two rods are charged, the ball goes back and forth between the two rods,
but once one rod is removed, the ball goes to one rod.