1. Exam 2: Review Session CH 24–27.3 & HW 05–07
Physics2113
Jonathan Dowling
Lecture 22: WED 14 OCT
Exam 2: Review Session
CH 24–27.3 & HW 05–07
Some links on exam stress:

2. Exam 2 (Ch 23) Gauss’s Law: Flux, Sphere, Plane, Cylinder
(Ch24) Sec.11 (Electric Potential Energy of a System of Point Charges); Sec.12 (Potential of Charged Isolated Conductor)
(Ch 25) Capacitors: capacitance and capacitors; caps in parallel and in series, dielectrics; energy, field and potential in capacitors.
(Ch 26) Current and Resistance. Current Density. Ohm’s Law. Power in a Resistor.

3. Gauss’ law
At each point on the surface of the cube shown in Fig , the electric field is in the z direction. The length of each edge of the cube is 2.3 m. On the top surface of the cube E = -38 k N/C, and on the bottom face of the cube E = +11 k N/C. Determine the net charge contained within the cube. [-2.29e-09] C

4. Gauss’s Law: Cylinder, Plane, Sphere

5. Problem: Gauss’ Law to Find E

6. Two Insulating Sheets

7. E does not pass through a conductor
Two Conducting Sheets
E does not pass through a conductor
Formula for E different by Factor of 2
7.68
4.86
4.86
7.68
12.54

8. Electric Fields With Insulating Sphere

10. Electric potential:
What is the potential produced by a system of charges? (Several point charges, or a continuous distribution)
Electric field lines, equipotential surfaces: lines go from +ve to –ve charges; lines are perpendicular to equipotentials; lines (and equipotentials) never cross each other…
Electric potential, work and potential energy: work to bring a charge somewhere is W = –qV (signs!). Potential energy of a system = negative work done to build it.
Conductors: field and potential inside conductors, and on the surface.
Shell theorem: systems with spherical symmetry can be thought of as a single point charge (but how much charge?)
Symmetry, and “infinite” systems.

11. Electric potential, electric potential energy, work
In Fig , point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles?

12. Derive an expression in terms of q2/a for the work required to set up the four-charge configuration of Fig , assuming the charges are initially infinitely far apart.
The electric potential at points in an xy plane is given by V = (2.0 V/m2)x2 - (4.0 V/m2)y2. What are the magnitude and direction of the electric field at point (3.0 m, 3.0 m)?

13. Potential Energy of A System of Charges
4 point charges (each +Q) are connected by strings, forming a square of side L
If all four strings suddenly snap, what is the kinetic energy of each charge when they are very far apart?
Use conservation of energy:
Final kinetic energy of all four charges = initial potential energy stored = energy required to assemble the system of charges +Q +Q +Q +Q
Do this from scratch! Don’t memorize the formula in the book!
We will change the numbers!!!

14. Potential Energy of A System of Charges: Solution+Q +Q
No energy needed to bring in first charge: U1=0
Energy needed to bring in 2nd charge: +Q +Q
Energy needed to bring in 3rd charge =
Total potential energy is sum of all the individual terms shown on left hand side =
Energy needed to bring in 4th charge =
So, final kinetic energy of each charge =

15. Potential V of Continuous Charge Distributions
Straight Line Charge: dq=λdx
λ=Q/L
Curved Line Charge: dq=λds
λ=Q/2πR
Surface Charge: dq=σdA
σ=Q/πR2
dA=2πR’dR’

16. Potential V of Continuous Charge Distributions
Curved Line Charge: dq=λds
λ=Q/2πR
Straight Line Charge: dq=λdx
λ=Q/L

17. Potential V of Continuous Charge Distributions
Surface Charge: dq=σdA
σ=Q/πR2
dA=2πR’dR‘
Straight Line Charge: dq=λdx
λ=bx is given to you.

18. ✔ ✔ ✔ ✔

19. ICPP: (a) True (b) False (c) True (d) True
Consider a positive and a negative charge, freely moving in a uniform electric field. True or false?
Positive charge moves to points with lower potential voltage.
Negative charge moves to points with lower potential voltage.
Positive charge moves to a lower potential energy.
Negative charge moves to a lower potential energy.
– – – – – – – –
(a) True –Q +Q +V –V
(b) False
(c) True
(d) True

20. electron ✔ ✔ ✔ ✔

21. downhill

22. No vectors!
Just add with sign.
One over distance.
Since all charges same and all distances same all potentials same.

24. (a) Since Δx is the same, only |ΔV| matters!
|ΔV1| =200, |ΔV2| =220, |ΔV3| =200
|E2| > |E3| = |E1|
The bigger the voltage drop the stronger the field. Δx
(b) = 3
(c) F = qE = ma
accelerate leftward

25. Capacitors E = σ/ε0 = q/Aε0 E = V d q = C V Cplate = ε0A/d
Connected to Battery: V=Constant
Disconnected: Q=Constant
Csphere=ε0ab/(b-a)

26. Isolated Parallel Plate Capacitor: ICPP
A parallel plate capacitor of capacitance C is charged using a battery.
Charge = Q, potential voltage difference = V.
Battery is then disconnected.
If the plate separation is INCREASED, does the capacitance C:
(a) Increase?
(b) Remain the same?
(c) Decrease?
If the plate separation is INCREASED, does the Voltage V: +Q –Q
Q is fixed!
d increases!
C decreases (= ε0A/d)
V=Q/C; V increases.

27. Parallel Plate Capacitor & Battery: ICPP
A parallel plate capacitor of capacitance C is charged using a battery.
Charge = Q, potential difference = V.
Plate separation is INCREASED while battery remains connected. +Q –Q
V is fixed constant by battery!
C decreases (=ε0A/d)
Q=CV; Q decreases
E = σ/ε0 = Q/ε0A decreases
Does the Electric Field Inside:
(a) Increase?
(b) Remain the Same?
(c) Decrease?
Battery does work on capacitor to maintain constant V!

28. Capacitors Capacitors Q=CV In series: same charge 1/Ceq= ∑1/Cj
In parallel: same voltage
Ceq=∑Cj

29. Capacitors in Series and in Parallel
What’s the equivalent capacitance?
What’s the charge in each capacitor?
What’s the potential across each capacitor?
What’s the charge delivered by the battery?

32. Capacitors: Checkpoints, Questions

33. Parallel plates: C = ε0 A/d Spherical:
Cylindrical: C = 2πε0 L/ln(b/a)]
Hooked to battery
V is constant.
Q=VC
(a) d increases -> C decreases -> Q decreases
(b) a inc -> separation d=b-a dec. -> C inc. -> Q increases
(c) b increases -> separation d=b-a inc.-> C dec. -> Q decreases

34. PARALLEL:
V is same for all capacitors
Total charge = sum of Q
SERIES:
Q is same for all capacitors
Total potential difference =
sum of V
Parallel: Voltage is same V on each but charge is q/2 on each since
q/2+q/2=q.
(b) Series: Charge is same q on each but voltage is V/2 on each since V/2+V/2=V.

35. Example: Battery Connected —
Voltage V is Constant but Charge Q Changes
Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E;
Battery remains connected
V is FIXED; Vnew = V (same)
Cnew = κC (increases)
Qnew = (κC)V = κQ (increases).
Since Vnew = V, Enew = V/d=E (same)
dielectric
slab
Energy stored? u=ε0E2/2 => u=κε0E2/2 = εE2/2
increases

36. Example: Battery Disconnected —
Voltage V Changes but Charge Q is Constant
Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E;
Battery remains disconnected
Q is FIXED; Qnew = Q (same)
Cnew = κC (increases)
Vnew = Q/Cnew = Q/(κC) (decreases).
Since Vnew < V, Enew = Vnew/d = E/κ (decreases)
dielectric
slab
Energy stored?

37. Current and Resistance
i = dq/dt
R = ρL/A
Junction rule
V = i R
E = J ρ
r = ρ0(1+a(T-T0))

38. Current and Resistance

39. A cylindrical resistor of radius 5. 0mm and length 2
A cylindrical resistor of radius 5.0mm and length 2.0 cm is made of a material that has a resistivity of 3.5x10-5 Ωm. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W?

41. Current and Resistance: Checkpoints, Questions

42. Fill in the blanks. Think water in hose!
26.2: Electric Current, Conservation of Charge, and Direction of Current:
Fill in the blanks.
Think water in hose!

43. All quantities defined in terms of + charge movement!
(a) right
(b) right
(d) right
(c) right

44. DC Circuits
Loop rule
Multiloop
Single loop
V = iR
P = iV

45. Resistors vs Capacitors
Resistors Capacitors
Key formula: V=iR Q=CV
In series: same current same charge
Req=∑Rj /Ceq= ∑1/Cj
In parallel: same voltage same voltage
1/Req= ∑1/Rj Ceq=∑Cj

46. Resistors in Series and in Parallel
What’s the equivalent resistance?
What’s the current in each resistor?
What’s the potential across each resistor?
What’s the current delivered by the battery?
What’s the power dissipated by each resisitor?

47. Series and Parallel

48. Series and Parallel

49. Series and Parallel

50. Problem: 27.P.018. [406649]
Figure shows five 5.00 resistors.
(Hint: For each pair of points, imagine that a battery is connected across the pair.)
Fig
(a) Find the equivalent resistance between points F and H.
(b) Find the equivalent resistance between points F and G.
Slide Rules:
You may bend the wires but not break them.
You may slide any circuit element along a wire so long as you don’t slide it past a three (or more) point junction or another circuit element.

53. Too Many Batteries!

54. Circuits: Checkpoints, Questions

55. +E
-iR
(a) Rightward (EMF is in direction of current)
(b) All tie (no junctions so current is conserved)
(c) b, then a and c tie (Voltage is highest near battery +)
(d) b, then a and c tie (U=qV and assume q is +)

56. (a) all tie (current is the same in series)
The voltage drop is –iR proportional to R since i is same.

57. (a) Vbatt < 12V (walking with current voltage drop –ir
(b) Vbatt > 12V (walking against current voltage increase +ir
(c) Vbatt = 12V (no current and so ir=0)