Physics 112 Formulae
Fundamental Constants
ρair (density of air) = 1.2 kg/m (20oC, 1 atm);
e = 1.602176487 X 10-19 C
me = 9.109382 X 10-31 kg
mp = 1.672621 X 10-27 kg
mn = 1.674927 X 10-27 kg
3
speed of sound in air = 344 m/s (20oC, 1 atm)
k = 1/4πεo = 8.987551787 X 109 N m2/C2
εo = 8.854187817 X 10-12 C2/Nm2
μo = 4π X 10-7 T m/A
1
𝑐 = 𝜀 𝜇 = 299,792,458 𝑚/𝑠 (exact)
√ 𝑜 𝑜
UNIT 1: Charges, Coulomb’s Law, Gauss’s Law
⃗ 12 | =
Coulomb’s Law: |𝑭
𝑘 𝑞1 𝑞2
1
=
𝑟12 2
⃗ |=
Point: |𝑬
⃗⃗ | =
Line: |𝑬
𝑟2
2𝑘𝜆
𝑟
⃗ | = 2𝜋𝑘𝜎 = 𝜎 or
Plane: |𝑬
2𝜀
𝑜
⃗ ∙ 𝑑𝑨
⃗⃗ = ∫ 𝐸⏊ 𝑑𝐴 = ∫ 𝐸 cos 𝜑 𝑑𝐴;
𝛷𝑬 = ∫ ⃗𝑬
4
Sphere: 𝑎𝑟𝑒𝑎 = 4𝜋𝑟 2 ;
𝑣𝑜𝑙𝑢𝑚𝑒 = 3 𝜋𝑟 3 ;
𝑞1 𝑞2
= 4𝜋𝜀
𝑟2
𝜆
𝑜
1
= 2𝜋𝜀
𝜎
𝜀𝑜
⃗𝑭 = 𝑞𝑬
⃗
;
4𝜋𝜀𝑜 𝑟12 2
𝑘𝑞
1
𝑞
𝑟
𝑜
if conducting plane
⃗ = 𝑞𝑒𝑛𝑐𝑙
Gauss’s Law: ∮ ⃗𝑬 ∙ 𝑑𝑨
𝜀
𝑜
Cylinder: 𝑎𝑟𝑒𝑎 = 2𝜋𝑟𝐿 + 2𝜋𝑟 2 ; 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝜋𝑟 2 𝐿
UNIT 2: Potential Energy, Voltage, Capacitors
𝑃𝐸 = 𝑈 =
𝑘𝑞0 𝑞1
𝑟01
𝑞
1
= 4𝜋𝜀
𝑞
𝑞
∆𝑃𝐸 = 𝑞𝑜 ∆𝑉 or 𝑃𝐸 = 𝑞𝑜 𝑉
𝑘𝑞0
1
,
𝑉
=
=
𝑟
𝑟
4𝜋𝜀
𝑞0 𝑞1
𝑜
01
𝑞3
1
𝑞
0
𝑞0
𝑟0
𝑜
𝑞3
𝑞
(𝑉 → 0 𝑎𝑠 𝑟 → ∞);
𝑃𝐸 = 𝑈 = 4𝜋𝜀0 ( 𝑟1 + 𝑟2 + 𝑟 + ⋯ ), 𝑉 = 4𝜋𝜀 ( 𝑟1 + 𝑟2 + 𝑟 + ⋯ ) (𝑉 → 0 𝑎𝑠 𝑟 → ∞);
𝑞
𝑜
1
𝑏 𝑑𝑞
𝑃𝐸 = 4𝜋𝜀0 ∫𝑎
𝑜
2
3
𝑜
1
2
3
𝑓 𝑑𝑞
1
𝑉 = 4𝜋𝜀 ∫𝑖
,
𝑟
𝑏
𝑏
𝑃𝐸𝑏 − 𝑃𝐸𝑎 = − ∫𝑎 ⃗𝑭 ∙ 𝑑𝒍 = − ∫𝑎 𝑞0 ⃗𝑬 ∙ 𝑑𝑙 ;
Infinite line charge: 𝑉𝑏 − 𝑉𝑎 =
𝜎
𝑟
𝑜
(𝑉 → 0 𝑎𝑠 𝑟 → ∞);
𝑏
𝑉𝑏 − 𝑉𝑎 = − ∫𝑎 ⃗𝑬 ∙ 𝑑𝒍;
𝑟
𝜆 ln( 𝑏 )
𝑟𝑎
2𝜋𝜀0
;
⃗ |𝑑)
(𝑉 = |𝑬
Infinite sheet: 𝑉𝑏 − 𝑉𝑎 = 2𝜀 (𝑟𝑏 − 𝑟𝑎 );
𝑜
𝑄 = 𝐶𝑉;
𝐶𝑒𝑞
for parallel plates, 𝐶 =
1
𝑄2 1
𝑃𝐸 = 𝐶𝑉 2 =
= 𝑄𝑉;
2
2𝐶 2
1
1
1
= 𝐶1 + 𝐶2 + ⋯;
= 𝐶 + 𝐶 + ⋯;
𝐶
1
2
⃗| ;
𝑢 = 𝜀𝑜 |𝑬
2
𝑒𝑞
𝐶𝑛𝑒𝑤 = 𝐾𝐶𝑜𝑙𝑑 ;
1
2
1
2
⃗| ;
𝑢 = 𝐾𝜀𝑜 |𝑬
2
𝐴𝜀𝑜
𝑑
;
UNIT 3: Current, Resistance, Introduction to Circuits
𝐼=
𝑑𝑄
𝐼
⃗ 𝒅;
𝑱 = 𝑛𝑞𝒗
= 𝑛|𝑞|𝑣𝑑 𝐴;
|𝑱| = 𝐴 = 𝑛|𝑞|𝑣𝑑 ;
𝑑𝑡
⃗ = 𝜌𝑱 and 𝑉 = 𝐼𝑅;
Ohm’s Law: 𝑬
𝑃 = 𝐼𝑉 = 𝐼 2 𝑅 =
1
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 + ⋯;
∑ 𝐼𝑖𝑛 = ∑ 𝐼𝑜𝑢𝑡 ;
𝑅𝑒𝑞
𝑉2
𝑅
𝑅=
𝜌𝐿
𝐴
;
;
1
1
1
= 𝑅 + 𝑅 + 𝑅 + ⋯;
1
2
3
∑𝑐𝑙𝑜𝑠𝑒𝑑 𝑙𝑜𝑜𝑝 ∆𝑉 = 0;
𝑡
𝑡
Charging: 𝑄(𝑡) = 𝑄𝑓 (1 − 𝑒 −𝑅𝐶 ) = 𝐶𝑉𝑜 (1 − 𝑒 −𝑅𝐶 ) ; 𝐼(𝑡) =
Discharging: 𝑄(𝑡) = 𝑄𝑜 𝑒
−
𝑡
𝑅𝐶
;
𝑄𝑜
𝐼(𝑡) = 𝑅𝐶 𝑒
−
𝑡
𝑅𝐶
;
𝑉𝑜
𝑅
𝑡
𝑒 −𝑅𝐶 ;
Physics 112 – Test #3
23 March 2015
Name _____________________________
Directions: This is a CLOSED BOOK test, and you may only use the calculator and cheat
sheet provided. SHOW ALL OF YOUR REASONING! You can only get partial credit
for an incorrect answer if you show your reasoning. You may have as much time as you
like to finish this exam.
1. (5 pts) When this unit began, I said we were leaving the realm of electrostatics. If there
is a current flowing through a wire that has resistance, the electric field inside of it (in
the “meat” of it) is NOT zero. Explain why this must be true.
2. (5 pts) The diagram at right shows 6 wires
coming together to form a junction. The currents
through 5 of the wires are as shown. Determine
the direction and magnitude of the current in the
wire with the ?? by it.
3A
6A
2A
4A
5A
??
3. A voltage of 3.0 volts is maintained across two cylinders
connected together in series as shown. Each cylinder
has a cross-sectional area of 2.0  10-5 m2 and length
0.005 m. The left cylinder has 5.0  1021 conduction
electrons per cubic meter and a resistivity of 1.0  10-2
Ω·m. The right cylinder has has 5.0  1022 conduction
electrons per cubic meter and a resistivity of 4.0  10-3
Ω·m. You may assume that the electric field inside each
cylinder is uniform. Determine
a) (7 pts) the resistance of each cylinder.
b) (6 pts) the current out of the battery.
c) (7 pts) the magnitude of the electric field inside each cylinder.
3.0 V
4. (15 pts) Rank the brightness of the bulbs in the below circuit from brightest to dimmest.
All bulbs are identical, and the wires connecting them have no resistance. The voltage
source is ideal. Explain your reasoning.
C
A
B
D
E
F
5. (18 pts) Consider the circuit below. All voltage sources are ideal (no internal
resistance) and the wires have negligible resistance. The current through battery V x is
zero. Determine
a) the voltage Vx and its orientation (which way the positive terminal points—is it
correct in the diagram or does it point the other way?), and
b) The current out of the 10 V battery (magnitude and direction).
15.0 Ω
4.0 Ω
27.0 V
7.0 Ω
15.0 V
26.0 V
9.0 Ω
5.0 Ω
11.0 Ω
10.0 V
8.0 Ω
12.0 V
Vx
12.0 Ω
6. (20 pts) A really horrible non-ideal voltmeter
(internal resistance Rv = 100 Ω) is connected
across the terminals of a non-ideal battery with
r = 5Ω
an internal resistance of 5.0 Ω. The voltmeter
100
Vo = ?
reads 7 volts.
Now the same battery is connected with an
ammeter (internal resistance RA = 5.0 Ω, a
100
resistor (R = 19 Ω), and this same
horrible voltmeter as shown at right.
a) What does the ammeter read?
b) What voltage does the horrible
r = 5Ω
voltmeter read now?
Vo = ?
V
RV = 100Ω
100
reads
volts
7
RA = 5Ω
A
R=
100
19
Ω
V
RV = 100Ω
7. Initially, the capacitor in the circuit at right is uncharged and the
switch is in position 2. The switch is put into position 1 so that
the capacitor begins to charge. After the switch has been in
position 1 for 7 minutes, the switch is moved back to position
2 so that the capacitor begins to discharge.
a) (10 points) Compute the voltages across the resistor and
across the capacitor just after the switch is thrown from
position 1 to position 2.
b) (7 points) Compute the charge on the capacitor 10.0 ms
after the switch is thrown from position 1 to position 2.
switch in
position 1
switch in
position 2
16.0 F
10.0 V
(ideal)
20 Ω
BONUS BRAIN BUSTER!!! (+5 points)
Reconsider the situation in problem number 3. Determine the total amount of charge (sign
and magnitude) sitting on the interface between the two cylinders. If you could not get 3c,
⃗⃗ | is 3000 N/C in the left cylinder and 1000 N/C in the right.
you may assume that |𝑬
Otherwise, use your solution for 3c.
3.0 V
Credit will only be given for correct reasoning. If you are able to guess the correct answer but
give no correct reasoning, you will receive no extra credit. However, if you give some correct
reasoning that would (or does) lead to the correct answer, at least some credit will be given.
________________________________________________
By signing my name above, I affirm that this test represents my work only, without aid
from outside sources. In all aspects of this course I perform with honor and integrity.