general physics ph 222-3a key - university of …people.cas.uab.edu/~mirov/test 1 spring 2011...
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GENERAL PHYSICS PH 222-3A (Dr. S. Mirov) Test 1 (02/02/11)
STUDENT NAME: ________________________ STUDENT id #: ___________________________ -------------------------------------------------------------------------------------------------------------------------------------------
key
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. 1 22
m mF G rr
Newton’s Law of Gravitation
2. 1 2 1 22 2
0
14
q q q qF r k rr r
Coulomb’s Law
3. 1 2 1 22 2
0
14
q q q qF k
r r , ε0=8.85x10-12 C2/Nm2 , k=8.99x109 Nm2/C2
4. 1C=(1A)(1s)
d q
5. d qid t
Electrical Current
Chapter 22
6. 0
FEq
Electric field, q0 is a positive test charge; F is electrostatic force that acts on
the test chargethe test charge
7. F q E
, Electrostatic force that acts on the particle with charge q.
8. 2 20
14
q qE r k rr r
, Electric field due to Point charge
1
9. 3 30 0
1 12 2
q d pEz z
, Electric field on axis z due to an electric dipole.
p=qd is an electric dipole moment with direction from negative to positive ends of a dipole.
1. 3 / 22 2
0
14
q zEz R
, Electric field due to charged ring
2. 20
14
qEz
, Electric field due to charged ring at large distance (z>>R) 0
3. 2 2
0
12
zEz R
, Electric field due to charged disk
4. 02
E
, Electric field due to infinite sheet
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
7. E A
, electric flux through a surface
8. E d A
, Electric flux through a Gaussian surface.
9. 0 e n cq , Gauss’ Law
10. 0 e n cE d A q
, Gauss’ Law
11. 0
E
, Conducting surface
12. 02
Er
, Line of charge
13. 0
E
, Electric field between two conducting plates
14. 20
14
qEr
, Spherical shell field at r≥R
E=0, Spherical shell field at r<R 2
1. 304
qE rR
, Uniform spherical charge distribution at r≤R
2. 20
14
qEr
, Uniform spherical charge distribution at r>R
Chapter 24
3. f iU U U W , 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. WUV
q q , Electric potential
6. f if i
U U U WV V Vq q q q
, Electric potential difference
7. f
i
V E d s
, Potential at any point f in the electric field relative to the zero potential at
point i. p
8. 0
14
qVr
, Potential due to point charge.
9. 0
14
n ni
ii i i
qV Vr
, Potential due to group of n point charges. 0i i i
10. 0
14
d qV d Vr
, Potential due to continuous charge distribution.
11. 1 / 22 2
l n4
L L dV
d
, Potential due to line of charge
04 d
f g
12. 2 2
02V z R z
, Potential due to charged disk
3
1. 20
1 ( )4
pCosVr
, Potential due to electric dipole.
2. ˆˆ ˆ
x y zE E i E j E k
, Calculating electric field from the potential
xVEx
, y
VEy
, z
VEz
y
3. VEs
, Calculating electric field from the potential in a simple case of uniform
electric field.f
4. 1 3 2 31 2
0 12 0 13 0 23
1 1 14 4 4
q q q qq qUr r r
, Potential energy of system of three
chargescharges.
5. , 10
14
ni j
i j iji j
q qU
r
, Potential energy of system of n charges
j
6.
7. 1
1 ( )2
1
n
other ii
n
U V i q
q
Alternative expression for Potential energy of system of n charges
10
1( )4
nj
otherj jij i
qV i
r
p f gy f y f g
4
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 Nm, what must be the dipole moment?
F+
p E
79
in scalar form sin(2.5 10 ) 2 0 10
p EpE
N m C
F‐
i
9( ) 2.0 10sin (300 / )sin 25
p C mE N C
x‐axis
9
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 of the electric field at x = R/2 on the x axis?g
x
R
ESEP
3 2
At x= / 2 charged insulating sphere generates electric field
Q= directed to the positive direction of x axis4 2 8S
R
Q RER R
4 2 8
A particle with a positive charge Q placed at a ditance 3 / 2 from x= /o oR R
R R
2generates electric field directed to the negative direction of x axis with a magnitude
Q QE 2 29342
The resultant electric field at x= / 2 is directed to the positive direction of x axis
po
o
ERR
R
Q Q Q
10
an 2 2 2
Qd has magnitude of 8 9 72o o o
Q QER R R
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 a conducting wire, what are the charges on the smaller and larger spheres?y g g g p
a) After connecting two spheres by a conducting wire some some charge froma) After connecting two spheres by a conducting wire some some charge fromsphere 1 will be transfered to sphere 2. This transfer will proceed untillpotentials of sphere 2 will be equal to potential 1. This translates into
1 2 21 (1)
4 4 2 2o o
Q Q QQR R
1 2
b) Of course total charge should be conserved (2)Q Q Q
23) Substituting (1) into (2) gives 2
2
Qc Q
Q
Q
11
2 12 ; 3QQ Q
3Q
8. What is the magnitude of the electric field at the point ˆˆ ˆ(3.00 2.00 4.00 )i j k m if the electric potential is given by V=2.00xyz2, where V is in volts and x, y, and z are inmeters?
W lWe apply 2
2
2.00
2 00
xVE yzxVE xz
2.00
4.00
y
z
E xzyVE xyzz
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
2 2 2 150V m 150 N C.x y zE E E E
12