Download Physics 231 Course Review, Part 2

9. Equipotential Surfaces
Equipotential Surfaces
are always perpendicular
to Electric Field Lines
The surface of a conductor is ALWAYS an Equipotential surface
10. Capacitance
C = Q/V
Capacitance has units of “farads”
1 farad = 1 Coulomb/Volt
Q
A
C=
=ε
0
Vab
d
Recall result from infinite flat plate
Integral is easy because field is uniform
Capacitance between two parallel plates”
a
r
b r
Va − Vb = ∫ E ⋅ dl = Ed
1 Qd
σ
Q
=
E=
ε0 ε0 A
V = Ed =
ab
ε0 A
+Q
-Q
To add more charge, I must do Work
Energy Storage in a Capacitor
Q
V=
C
To add a small amount of charge dq I must do work dW -
q
dW = Vdq = dq
C
Q final
Q final
Qinitial
Q final
∫ dW = ∫ Vdq = ∫
Qinitial
Stored Energy in a Charged Capacitor
1 2 ⎞
1 ⎛1
1
q
qdq = ⎜ Q 2final = Qinitial
dq =
⎟
∫
2
C⎝2
C Qinitial
C
⎠
Q final
To add a finite charge Q’ I must integrate -
W=
Qinitial
If I start from NO net charge I get -
1
1 Q2 1
W=
= QV = CV 2
2 C 2
2
Capacitors in Parallel
Q2 = C2V
Q1 = C1V
Q = Q1 + Q2 = (C1 + C2 )V
Q
= C1 + C2
V
Cequiv = C1 + C2
Cequiv = C1 + C2 + C3 + ....
Capacitors in Series
Vac = V1 =
Q
C1
Vcb = V2 =
⎛ 1
1 ⎞
Vac = V = V1 + V2 = Q⎜⎜ + ⎟⎟
⎝ C1 C2 ⎠
V
1
1
=
+
Q C1 C2
1
Cequiv
1
Cequiv
=
1
1
+
C1 C2
=
1
1
1
+
+
+ ....
C1 C2 C3
Q
C2
11. Resistance and Ohms Law
Definition:
A Resistor is a device that follows Ohms Law:
V=IR
If there is a voltage “across” a resistor
then
There MUST be a current “through” it.
If there is a current “through” a resistor
then
There must be a voltage across it.
Step 2- Label ALL Currents – Use Kirchhoff’s Junction Rule to simplify
The directions of currents you chose is NOT IMPORTANT
If you guess wrong – Mr. Kirchhoff will give you a minus
sign for the current.
Step 5 – Go around the loops to get the equationsNOTE – If we go from + to – potential change is NEGATIVE
If we go from - to + potential change is POSITIVE
+
+
+ +
+
13V −
13V − I 1 (1Ω ) − (I 1 − I 3 )(1Ω ) = 0
− I 2 (1Ω ) − (I 2 + I 3 )(2Ω ) + 13V = 0
− I 1 (1Ω ) − I 3 (1Ω ) + I 2 (1Ω ) = 0
Loop 1
Loop 2
Loop 3
Step 6- Solve the Simultaneous Equations
13V − I 1 (1Ω ) − (I 1 − I 3 )(1Ω ) = 0
Loop 1
− I 2 (1Ω ) − (I 2 + I 3 )(2Ω ) + 13V = 0
Loop 2
− I 1 (1Ω ) − I 3 (1Ω ) + I 2 (1Ω ) = 0
Loop 3
Solve Equation 3 for I2:
I 2 = I1 + I 3
Plug into Equations 1 and 2 (this gives 2 equations and 2 unknowns):
13V = I 1 (2Ω ) − I 3 (1Ω )
Note Y&F have misprint here p991
13V = I 1 (3Ω ) + I 3 (5Ω )
Multiply top equation by 5 and add two equations:
65V = I 1 (10 Ω ) − I 3 (5Ω )
13V = I 1 (3Ω ) + I 3 (5Ω )
78V = I 1 (13Ω )
I1 = 6 A
I 3 = −1 A
Negative sign here means we
guessed wrong about direction of I3
I2 = 5 A