PH 211 07-24-2014

Physics 211 Exam-2

NAME: _____________________________________ Write down your name also on the back of your exam.

SIGNATURE and ID: ___________________________ Write down your name on the back of the package of sheets you turn in.

Notice: The exam is worth 110 points: Q1= 30 points, Q2= 40 points, and Q3= 40 points. (i.e. you have an opportunity to earn extra 10 point in this exam) .

1. 1A The figure at the right shows portions of two infinitely large, parallel, non-

conducting sheets, each with a fixed uniform charge. The surface charge

densities are 1 = 6 C/m2 and 2 = - 2 C/m2.

+

S

+

+

+

+

+

+

+

+

+

-

-

-

-

-

-

P U

(7.5 points) Three points P, S, and U are located at the vertices of a equilateral

triangle of side 10 mm. Rank the points according to the electrical potential at those positions respectively, greatest first. (Circle your answer).

a) U, P, S b) P, U, S c) P, S, U

d) U, S, P e) NA

(7.5 points) Three electrons are located at the vertices of a equilateral

triangle PSU of side 10 mm. Rank the electrons according to their electrical potential energy, greatest first. (Circle your answer).

a) U, P, S b) P, U, S c) P, S, U

d) U, S, P e) NA

1B (7.5 points) The figure shows three situations in which a positively charged

particle moves at velocity v

through a uniform magnetic field B

and experiences a magnetic force. In each situation, determine whether the orientation of the vectors are physically reasonable. (Circle your answer).

FB

v

B

x

FB

v B .

FB

v

B

.

I II III

a) F, F, T b) T, F, T c) T, T, F d) T, F, F e) NA

1C (7.5 points) The figure shows four directions for the velocity vector v

of a

negatively charged particle moving through a uniform magnetic field B

pointing to the left. (Of course, the magnetic fields exist all over the space) . Rank the directions according to the magnitude of the net force on the particle, greatest first. (Circle your answer).

1

3

-

2

4

B

a) 1,2,3,4 b) 1 tie with 4, 2, 3 d) 2, 3 tie with 4, 1

c) 3, 2, 1 tie with 4 e) NA

2. A total charge of Q = 4 x 10-6 Coulomb is uniformly distributed all over the volume of a

non-conducting sphere of radius R= 3 cm. The electric field is radially directed and has a magnitude equal to:

rR

rE3

0

Q

4

1)(

for r < R

2

0

Q

4

1)(

rrE

for r > R

The potential is chosen to be zero at r=R, i.e. V(R) = 0 Volts:

2A (12 points) Find general expression for the potential V(r), for the range r <R . (i.e. give an expression for the potential V as a function of r that satisfies the

condition V(R) = 0 Volts ). Note: Provide all your detailed procedure leading to your answer. Writing

directly just the answer will receive zero points in the grading).

2B (12 points) Find general expression for the potential V(r), for the range r > R. (i.e. give an expression for the potential V as a function of r that satisfies the

condition V(R) = 0 Volts). Note: Provide all your detailed procedure leading to your answer. Writing

directly just the answer will receive zero points in the grading).

2C (12 points) Calculate the potential at r =0; i.e. calculate V(0). Calculate the potential at infinity, i.e. calculate V(∞).

2D (4 points) Sketch the potential V as a function of r, from r=0 to r = ∞.

Hint: Use the expression for the potential-difference f

iif rE

dVV , and

capitalize on the fact that you can choose arbitrarily what point constitutes the initial point and what constitutes the final point. I would suggest to choose the initial and final points in such a way that the

electric field E

and the displacement r

d point in the same direction.

To double check that the signs ( + or -) in the calculations above are correct, you may want to verify them by using the alternative definition of potential- difference” based on the work done by an external force:

0

)()()( 0

q

W qext BAAVBV

(where A and B can be chosen arbitrarily as

the initial or final positions, respectively)

3. The figure shows a parallel plate capacitor with a plate area A = 24 (mm)2 and a separation distance d = 1 mm. The left half of the gap is filled with material of dielectric constant K1=3.14 ; the right half is filled with material of dielectric constant K2= 6.28. The capacitor is connected to a 8 V battery.

Side view

d= 1 mm 8 Volts K K

3A (12 points) Calculate the electric field on the left-half of the capacitor. Calculate the electric field on the right-half of the capacitor

3B (4 points) Sketch a possible charge distribution on each plate.

3B (12 points)

Calculate the surface charge density on the left-half of the plates.

Calculate the surface charge density on the right-half of the plates.

3C (12 points) Calculate the capacitance of the parallel-plate capacitor.

Some formulas:

nano

ln( ab ) = ln( a ) + ln ( b ) ln( a / b ) = ln( a ) - ln ( b )

Electron mass: 9.1 x 10-31

Kg Proton mass = 1.67 x 10-27

Kg

Electron charge 1.6 x 10-19

C

1 Gauss = 10-4

Tesla

Centripetal acceleration: R

ac

2v

1eV = 1.6 x 10-19

J

o= 8.85 x 10-12 C

2/N.m

2 , o = 9 x 10

9 N.m2/C

2,

o/= 10-7 T m/A

sin,cossin dCosd , 2sin2

1 cossin d

ELECTRICITY

Coulomb's Law: rur

qqF ˆ

4

12

12

0

Electric filed for an infinite and uniformly-charged sheet: E = /2o

is the surface charge density.

For a uniformly charged ring with radius R and total charge q:

2/322

0 )(4

1)(

Rz

zqzE

(Field along the axis passing through the center of the ring)

Gauss' Law = S

sdE

= q /o , where q is the net charge inside the

Gaussian surface S

Definition of Electric Potential 0

)()( 0

q

rWrV

qext

Assuming the potential at infinity is zero

Definition of Electric Potential difference 0

)()()( 0

q

W qext ABBVAV

Electric potential due to a point charge q: r

qrV

04

1)(

(Assuming the potential at infinite is zero)

Potential difference : f

iif rE

dVV

Relationship between E and V: Ex = - dV/dx

About capacitance

Q = C V U = CV2 / 2 = Q2 / 2 C

For a capacitor of parallel plates: C = Ao /d

RC circuit: Time constant = RC

MAGNETISM

F = q v x B F = force, q= charge, v = velocity, B = magnetic field

Magnetic field produced by a charge q that moves with velocity v

rvB

x4 3

0

r

q

= L i = Magnetic flux, L = inductance, i = current

Hall effect BI = nqtVHall

Inductive reactance XL = L

B

0 I

4

1

R Magnetic field at the center of a semi-circle of radius "R"

B 0 I

4 R Magnetic field at the center of an arc of angle f (in radians) and

radius "R".

B 0 I

2 r Magnetic field produced by a infinitely long wire at a distance "r"

from it.

F

L 0

IaIb

2 d Force per unit length between two parallel long wires,

carrying currents Ia and Ib respectively, separated by a

distance "d"

Faraday's Law t

,

where = Magnetic flux and = electromotive force

Definition of the magnetic dipole moment of a loop of area A, carrying a current I:

= I A n

where A = area, I current, n = unit vector perpendicular to the loop