GENERAL PHYSICS PH 222-3A (Dr. S. Mirov) Test 2 (02/23/11)

STUDENT NAME: ______KEY_____________ STUDENT id #: ___________________________ -------------------------------------------------------------------------------------------------------------------------------------------

ALL QUESTIONS ARE WORTH 30 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 charge g

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. 2

14

qE , Electric field due to charged ring at large distance (z>>R)204 z

f g g g ( )

3. 2 2

0

12

zEz R

, Electric field due to charged disk

4. 02

E

, 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

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 0 e n c

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. 2

14

qEr

, Uniform spherical charge distribution at r>R 04 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

U

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.

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.

10. 0

14

d qV d Vr

, Potential due to continuous charge distribution.

11. 1 / 22 2

0

l n4

L L dV

d

, Potential due to line of charge

12. 2 2

02V z R z

, Potential due to charged disk

3

1. 20

1 ( )4

p C o sVr

, Potential due to electric dipole.

2. ˆˆ ˆ

x y zE E i E j E k

, Calculating electric field from the potential

V V Vx

VEx

, y

VEy

, z

VEz

3. VEs

, Calculating electric field from the potential in a simple case of uniform

electric field.

4. 1 3 2 31 2

0 1 2 0 1 3 0 2 3

1 1 14 4 4

q q q qq qUr r r

, Potential energy of system of three

charges.

5. , 10

14

ni j

i j i j

q qU

r

, Potential energy of system of n charges j

i j

6.

7. 1

10

1 ( )2

1( )4

n

o t h e r ii

nj

o t h e rj j ij i

U V i q

qV i

r

Alternative expression for Potential energy of system of n charges

j i

Chapter 25

8. Capacitance is defined from q C V ,

9. A parallel plate capacitor o ACd

10 A cylindrical capacitor 2 LC 10. A cylindrical capacitor 2l n ( )oC

b a

11. A spherical capacitor 4 oa bC

b a

12. Isolated sphere 4 oC R

13. n capacitors in parallel 1

n

e q jj

C C 1j

14. n capacitors in series 1

1 1n

je q jC C

15. The electric potential energy of a charged capacitor 2

212 2qU C V

C 4

1. The energy density in vacuum 212 ou E

2. Capacitance with a dielectric filling the space between the plates a i rC k C

3. Gauss law with a dielectric qk E d A

o

Chapter 26.

4. An electric current d qid t

5. Current and current density relationship i J d A

6. Current density relationship with a drift speed and carrier charge density ( ) dJ n e v

7. Resistance of a conductor VRi

8. Resistivity and conductivity of materials 1 E

J

; E J

9 Resistance of a conductive wireLR 9. Resistance of a conductive wire RA

10. Change of resistivity with temperature ( )o o oT T

11. Ohm’s Law VRi

is independent of the applied potential difference V.

12. Resistivity of metal 2

me n

13. Power P i V

14. Resistive dissipation 2

2 VP i RR

Chapter 27.

15. Emf d Wd q

d q

16. Loop rule: the algebraic sum of electromotive forces and potential changes across circuit elements in a complete traversal of any loop of a circuit must be zero

17. Single loop circuits iR r

18. Power 2; ; e m fP i V P i R P i

19 S i in

R R19. Series resistances 1

e q jj

R R

20. Parallel resistances 1

1 1n

je q jR R

5

h h (1 )t RC t RCdqC 1. RC circuits. Charging the capacitor (1 ); t RC t RCdqq C e i e

dt R

2. Discharging the capacitor ; t RC t RCoo

qdqq q e i edt RC

Chapter 28. Magnetic fields

3. Magnetic force exerted on a point charge by a magnetic field: BF qv B

Bi

4. The number density n of charge carriers (Hall Effect): Bin

Vle

5. Circular orbit in magnetic field: mvrq B

6. Magnetic force on a straight current carrying wire of length L: BF iL B

7. The force acting on a current element in a magnetic field: BdF idL B

8. Magnetic dipole moment of current loop: = [current] x [area] 9. Torque on a current carrying coil: B

10. Orientation energy of a magnetic dipole: ( )U B

11. The work done on the dipole by the agent is: a f iW U U U 12 P bilit t t 1 26 10-6 N 2/C2 1 26 10-6 H/

6

12. Permeability constant: o = 1.26x10 6 Ns2/C2 = 1.26x10 6 H/m; 13. Permittivity constant: o = 8.85x10-12 C2/(Nm2) = 8.85x10-12 F/m;

1.

7

2.

8

3.

9

4.

10

5.

11

6. The current density magnitude in a certain circular wire is J=(2.75x1010 A/m4)r2, where ris the radial distance out to the wire’s radius of 3.00 mm. The potential applied to thep ppwire (end to end) is 60.0 V. How much energy is converted to thermal energy in 1.00 h?

Assuming the current is along the wire (not radial) we find the current from:

i = | J

| dA = 2

02

Rkr rdr =

12 kR4 = 3.50 A

where k = 2.75 1010 A/m4 and R = 0.00300 m. The rate of thermal energy generation isfound from P = iV = 210 W. Assuming a steady rate, the thermal energy generated in 3600 s is Q P t (210 J/s)(3600 s) = 7.56 105 J.

12

7.

13

8.

6.9.

15