Download TEST 1 over Chapters 21

GENERAL PHYSICS PH 222-3A (Dr. S. Mirov)
Test 1 (02/02/11)
key
STUDENT NAME: ________________________ STUDENT id #: ___________________________
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ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.
NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)
Important Formulas:
Chapter 21
1.

m 1m
F  G
r2
2.

F 
3.
F 
1
4 
4 

q1q 2 
q q 
r  k 1 2 2 r Coulomb’s Law
2
r
r

q1 q 2
r2
0
1

r Newton’s Law of Gravitation
2
0
4.
1C=(1A)(1s)
5.
i 
 k
q1 q 2
, ε0=8.85x10-12 C2/Nm2 , k=8.99x109 Nm2/C2
r2
dq
Electrical Current
dt
Chapter 22
6.


F
E 
q0
Electric field, q0 is a positive test charge; F is electrostatic force that acts on
the test charge
7.


F  q E , Electrostatic force that acts on the particle with charge q.
8.

E 
9.
E 
1
4 
1
2 

q 
q 
r  k 2 r , Electric field due to Point charge
r2
r

qd
1

2 
z3
0
0

0
p
, Electric field on axis z due to an electric dipole.
z3
p=qd is an electric dipole moment with direction from negative to positive ends of a
dipole.
1
1.
2.
3.
4.
E 
E 
1
4 
0
1
4 

2

E 
2

0
z
0
qz
2
 R
2

3 / 2
, Electric field due to charged ring
q
, Electric field due to charged ring at large distance (z>>R)
z2

1 

E 


z
z
2
 R
2

 , Electric field due to charged disk

, Electric field due to infinite sheet
0


5.
  p  E  p E S i n (  ) , Torque on a dipole in electric field
6.
U


  p  E   p E C o s (  ) , Potential energy of a dipole in electric field.
Chapter 23


E   A , electric flux through a surface
7.
 

8.
 

9.
 0   q e n c , Gauss’ Law
10. 
0



E  d A , Electric flux through a Gaussian surface.


E  d A  q e n c , Gauss’ Law
11. E 

, Conducting surface
0
12. E 

, Line of charge
2  0r
13. E 

, Electric field between two conducting plates
0
14. E 
1
4 

0
q
, Spherical shell field at r≥R
r2
E=0, Spherical shell field at r<R
2
1.

q
E  
4


0 R

2.
E 
1
4 

0

 r , Uniform spherical charge distribution at r≤R

3
q
, Uniform spherical charge distribution at r>R
r2
Chapter 24
3.
U
 U
 U
f
 W
i
, Change in electric potential energy due to work done by
electrostatic force.
4.
U   W  , Potential energy; W∞ is work done by electrostatic force during particle
move from infinity.
5.
V 
6.
V  V
7.
V  
U
q
 
W 
, Electric potential
q
 Vi 
f
U
f

q
U i
U

q
q
 
W
q
, Electric potential difference


E  d s , Potential at any point f in the electric field relative to the zero potential at
f

i
point i.
p
8.
9.
V 
V 
1
4 
n


0
Vi 
i
10. V


dV 

11. V 
4 
12. V


2
0
0

q
, Potential due to point charge.
r
4 
1
4 

i
0
0
2
dq
, Potential due to continuous charge distribution.
r

 L 
ln 


z
qi
, Potential due to group of n point charges.
ri
n
1
 R
L
2
 d
d
2
 z
2

1/ 2

 , Potential due to line off charge
g


 , Potential due to charged disk
3
1. V 
1
4 0

pCos ( )
, Potential due to electric dipole.
r2

2. E  Ex iˆ  E y ˆj  Ez kˆ , Calculating electric field from the potential
Ex  
3.
E
V
V
V
, Ey  
, Ez  
x
z
y
V
, Calculating electric field from the potential in a simple case of uniform
s
electric field.
f
4. U 
1
4 0

q1q2
1 q1q3
1 q2 q3
, Potential energy of system of three




4 0 r13 4 0 r23
r12
charges.
charges
5. U 
1
4 0
n

qi q j
i , j 1
i j
rij
, Potential energy of system of n charges
6.
7.
1 n
U   Vother (i )qi
2 i 1
Vother (i ) 
1
4 0
n
p
ffor Potential energy
gy off system
y
off n charges
g
q j Alternative expression
r
j 1
j i
ji
4
5
6
7
4.
9.6x10^‐16 N
8
5. A uniform electric field of 300N/C makes an angle of 25 with the dipole moment
of an electric dipole. If the torque exerted by the field has a magnitude of 2.5x10-7
Nm, what must be the dipole moment?
F+
 
  p E

in scalar form   pE sin 

( 107 N  m)
(2.5
 p

 2.0
2 0 109 C  m
E sin  (300 N / C ) sin 25
F‐
x‐axis
i
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6.
Positive charge Q is distributed uniformly throughout an insulating sphere of radius R,
centered at the origin. A particle with positive charge Q is placed at x = 2R on the x axis. What is
the magnitude
g
of the electric field at x = R/2 on the x axis?
EP
ES
x
R
At x=R / 2 charged insulating sphere generates electric field
 Q R
Q
ES  
=
directed to the positive direction of x axis
3 
2
4

R
2
8

R
o
o


A particle with a positive charge Q placed at a ditance 3R / 2 from x=R / 2
generates electric field directed to the negative direction of x axis with a magnitude
Q
Q
Ep 

2
9 o R 2
 3R 
4 o 

 2 
The resultant electric field at x=R / 2 is directed to the positive direction of x axis
and has magnitude of E 
Q
8 o R 2

Q
9 o R 2

Q
72 o R 2
10
7.
Two conducting spheres are far apart. The smaller sphere carries a total charge Q. The
larger sphere has a radius that is twice that of the smaller and is neutral. After the two spheres are
connected by
y a conducting
g wire, what are the charges
g on the smaller and larger
g spheres?
p
a) After connecting two spheres by a conducting wire some some charge from
sphere 1 will be transfered to sphere 2. This transfer will proceed untill
potentials of sphere 2 will be equal to potential 1. This translates into
Q1
Q2
Q

 Q1  2
(1)
4 o R 4 o 2 R
2
b) Of course total charge should be conserved
Q1  Q2  Q
c) Substituting (1) into (2) gives
 Q2 
(2)
3Q2
Q
2
2Q
Q
; Q1 
3
3
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8. What is the magnitude of the electric field at the point
(3.00iˆ  2.00 ˆj  4.00kˆ) m if the
electric potential is given by V=2.00xyz2, where V is in volts and x, y, and z are in
meters?
W apply
We
l
V
 2.00 yz 2
x
V
Ey  
 2.00
2 00 xz 2
y
V
 4.00 xyz
Ez  
z
Ex  
which, at (x, y, z) = (3.00 m, –2.00 m, 4.00 m), gives
(Ex, Ey, Ez) = (64.0 V/m, –96.0 V/m, 96.0 V/m).
The magnitude of the field is therefore

E  Ex2  E y2  Ez2  150 V m  150 N C.
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