PHYSICS 231 - Solution Key to Sample Test 3
1a. From F = qv B , we have F = 0 if v is parellel to B . So q will experience no force if it
moves along a magnetic eld line.
1b. All turns in the coil carry the same current. Therefore, neighboring turns carry
Hannah Jarrett
Breanna Rhyne
Physics Lab 231
Section 006
Date of Experiment: 11/04/2013
RC-RL Circuits
Date of Submission: 11/11/2013
Objective(s):
The objectives of this experiment were to understand the time behavior of voltage and
currents as they flow
Hannah Jarrett
Breanna Rhyne
Physics Lab 231
Section 006
Date of Experiment: 11/11/2013
AC Circuits 1 & 2
Date of Submission: 11/25/2013
Objective(s):
The objectives of this experiment were to understand how current and voltage are
dependent upon time in
PHYSICS 231 - Solution Key to the Final Exam
1a. If you connect N bulbs each of resistance R in series to a battery of emf E , the current is
I=
E
NR
The power in each bulb is
E2
N 2R
As N increases, P decreases, so each bulb gets dimmer.
P = I 2R =
1b. T
PHYSICS 231 FINAL EXAM
NAME:
STUDENT ID #:
USEFUL CONSTANTS
0
= 8.85 1012 C 2 /(N m2 )
due to innite sheet:
E=
20
1
k=
= 8.988 109 N m2 /C 2 Potential energy:
4 0
q1 q2
U =k
e = 1.6 1019 C
r
me = 9.1 1031 kg
Potential:
7
0 = 4 10 T m/A
U
V=
g = 9.81 N/kg
PHYSICS 231 - Solution Key to the Sample Final Exam
1a. No. By Gausss Law,
E =
Qenc
0
regardless of the shape of the surface.
1b. Inserting the dielectric increases the capacitance (C = KC0 > C0 ). The voltage remains the
same, because we keep the capacit
PHYSICS 231 - Solution Key to Test 1
1a. The force is F = qE , so the proton and electron experience opposite forces (eE and eE ,
respectively).
From F = ma, we see that the acceleration of the proton is smaller than the acceleration of the
electron, beca
PHYSICS 231 - Solution Key to Test 3
1a. The energy is 1 mv 2 . It is doubled if v
2
so R v and R
2v . The radius is
mv
R=
|q |B
2R.
1b. All turns in the coil carry the same current. Therefore, neighboring turns carry currents in the
same direction and
PHYSICS 231 TEST # 3
NAME:
STUDENT ID #:
USEFUL CONSTANTS
0
= 8.85 1012 C 2 /(N m2 )
due to wire (Biot-Savart law):
0 I
4
B=
e = 1.6 1019 C
me = 9.1 1031 kg
Amp`res law:
e
0 = 4 107 T m/A
USEFUL FORMULAS
Magnetic force
on point charge: F = qv B
on wire: F
PHYSICS 231 - Solution Key to Test 2
1a. Before the metal is inserted, the capacitance is C = 0 A/d. After it is inserted, we have
effectively two capacitors connected in series of capacitances C1 = 0 A/d1 and C2 = 0 A/d2 ,
respectively. The capacitance i
PHYSICS 231 TEST # 2
NAME:
STUDENT ID #:
USEFUL CONSTANTS
0
Current and current density:
= 8.85 1012 C 2 /(N m2 )
1
k=
= 8.988 109 N m2 /C 2
4 0
e = 1.6 1019 C
me = 9.1 1031 kg
USEFUL FORMULAS
Capacitance: C = Q/V
Parallel-plate capacitor: C =
Capacitors
PHYSICS 231 SAMPLE TEST # 3
Problem 1
(a) Can a charged particle move in a magnetic eld without experiencing any
force? If so, how? If not, why not?
(b) A current was sent through a helical coil spring. The spring contracted, as
if it had been compressed.
Electric Fields Experiment Lab Report
Zhiheng Zhu
000435200
PHY 231
Lab Instructor: Mohammad Al-Mamun
Partner: Jordan Pruitt
8/30/2016
Argument
Electric field is defined as the electric force per unit charge. In this lab, the field can be detected
through