Lecture 7

Practice Problems

Prof. Viviana Vladutescu

Suppose D = 6rcos φ aφ C/m2.(a) Determine the charge density at the point (3m, 90, -2m). Find the total flux through the surface of a quartered-cylinder defined by 0 ≤ r ≤ 4m, 0 ≤ φ ≤ 90, and -4m ≤ z ≤ 0 by evaluating (b) the left side of the divergence theorem and (c) the right side of the divergence theorem.(a)

6 cos1 16sin .

cylinder

D

D

33,90 , 2 6 .v

C

m

(b)

0 90

,top bottom outside

d

D S

Divergence Theorem and Gauss’s Law

192 .d CD S

0

0

6 cos 192d dz C

a a

90

90

6 cos 0d dz

a a

note that the top, bottom and outside integrals yield zero since there is no component of D in the these dS directions.

So,

(c)

90 4 0

0 0 4

6sin ,

6 sin 192 .

dv d d dz

dv d d dz C

D

D

2 212 1212 24 x y x y

xy xy VV y xy

x y m

2E a a a a

22

-1.1 2.2 r o x y

nCy xy

m D E a a

The potential field in a material with εr= 10.2 is V = 12 xy2 (V). Find E, P and D.

Electric Potential

1 9.2e r

12 22

9.2 8.854 10 -9.8 2.00 e o x y

nCx y xy

m P E E = a a

For z ≤ 0, r1 = 9.0 and for z > 0, r2 = 4.0. If E1 makes

a 30 angle with a normal to the surface, what angle does E2 make with a normal to the surface?

1 1 1 2 2 2 1 2sin , sin , and T T T TE E E E E E

Boundary Conditions

1 1 1 1 2 2 2 2 1 2cos , cos , and since 0N r o N r o N N sD E D E D D

1 2

1 2

,T T

N N

E E

D D Therefore

and after routine math we find 1 22 1

1

tan tanr

r

Using this formula we obtain for this problem 2 = 14°.

also

1. A spherical capacitor consists of a inner conducting sphere of radius R1=3cm, and surface charge density ρs=2nC/m2 and an outer conductor with a spherical inner wall of radius R2=5cm . The space in between is filled with polyethylene.a) Determine the electric field intensity from ρs between plates(15 points)b) Determine the voltage V12 (10 points)c) Determine the capacitance C (10 points)

2. Consider a circular disk in the x-y plane of radius 5.0 cm. Suppose the charge density is a function of radius such that ρs = 12r nC/cm2 (when r is in cm). Find the electric field intensity a point 20.0 cm above the origin on the z-axis.(25 points)

3. Assume the z=0 plane separates two lossless dielectric regions, where the first one is air and the second one is glass. If we know that the electric field intensity in region 1 is E1 =2y ax -3x ay+(5+z) az, a)Find E2 and D2 at the boundary side of region 2? (20 points)b)Can we determine E2 and D2 at any point in region 2? Explain.(5 points)

4. Two spherical conductive shells of radius a and b (b > a) are separated by a material with conductivity σ. Find an expression for the resistance between the two spheres.(20 points)

For a ring of charge of radius a,

3

2 2 2.

2

L z

o

ah

a h

aE

Now we have L=sd and

3

2 2 2

,

2

z

o

A d hd

h

aE

where s = A nC/cm2.

Now the total field is given by the integral:

2

32 2 2

.2

z

o

Ah d

h

a

E

Problem 2

This can be solved using integration by parts, where u = , du = d,

2 2 2 2

1, and .

dv dv

h h

This leads to

2 2

2 2ln .

2 zo

Ah a a a h

ha h

E a

Plugging in the appropriate values we arrive at E = 6.7 kV/cm az.

Problem 3

zyx aaxayE 5321

)0()0(

32)0()0(

21

21

zDzD

axayzEzE

NN

yxTT

)5()0(

)0()0(

2

12

2211

zN

NN

azE

zEzE

At the z=0 plane

From the boundary condition of two dielectrics we get

022

12

2

12

32

32

rzyx

zyx

aaxayD

aaxayE

z=0

Problem 4- Hint

l

VElEV ab

ab

s S

IJJSdsJI

l

V

S

IEJ ab

l

J

σ Vab

a

b

I

)( where

S

lRI

S

lVab

Resistance of a straight piece of homogeneous material of a uniform cross section for steady current

s

b

aab

dsE

ldE

I

VR

Resistances connected in series

Resistances connected in parallel

nsr RRRRR ..........321

nRRRRR

1.............

1111

321//

Problem 4First find E for a < r < b, assuming +Q at r = a and

–Q at r = b. From Gauss’s law:24 r

o

Q

rE a

Now find Vab:

2

2

4

1 1 1.

4 4 4

a a

ab r rob b

aa

bo o ob

QV d dr

r

Q dr Q Q

r r a b

E L a a

Now can find I:

22

2

0 0

1sin

sin .

r ro

o o

QI d = d = r d d

4 r

Q Qd d

4

J S E S a a

Finally,

1 1 1

4abV

RI a b