GENERAL PHYSICS PH 222-3A (Dr. S. Mirov) Test 1 (02/02/11)

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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

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 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