PHYS 222 Worksheet 4 - Gauss’s Law and Conductors Supplemental Instruction Iowa State University

Leader: Alek Jerauld

Course: PHYS 222

Instructor: Dr. Paula Herrera-Siklódy

Date: 1/24/12

Useful Equations

The uniform electric field flux relative to the charge

The electric field flux through a Gaussian surface Related Problems 1) A sphere of radius a = 2.0 cm with a uniformly distributed charge Q1 = 4.0 μC is at the center of a conducting spherical shell of radii b = 4.0 cm and c = 6.0 cm that has a total charge Q2 = −8.0 μC. (a) Find the electric field at point A, which is at r = 1.0 cm from the center of the system.

(b) Find the electric field at point B, which is at r = 4.0 cm from the center of the system

0/E enclosedq

E E d A

0

3

1 13

1

0

2

2

61 1

2

0 0

/ 0.5 / 2

/

4( / 2)

3 / 84

( )3

8

42

4 45(10 ) /2 8 8

enc

enc

r a r a

q

a

q Q Q

a

Q

aE dA E

Q QaE E N C

a

0 1 0

2

2 611 0 2

0

/ /

4

4 / 22.5(10 ) /4

encq Q

E dA E b

QE b Q E N C

b

(c) Find the electric field at point D, which is at r = 8.0 cm from the center of the system

2) Negative charge, –Q, is distributed uniformly over the surface of a thin spherical shell with radius R. (a) Calculate the magnitude of the force the shell exerts on a positive charge q located a distance r > R from the center of the shell (outside the shell). (b) Calculate the magnitude of the force that the shell exerts on a positive point charge q located at a distance r < R from the center of the shell (inside the shell). Express your answers in terms of the variables q, Q, r, R, and constants π and Є0.

(a)

(b)

3) A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is -3Q, and it is insulated from its surroundings.

1 20

0

2

6

2 61 2 1 2

2 2

0 0 0

/

4

4 8 (10 )4 5.62(10 ) /

4 4 0.08

enc

D

D

D

Q Qq

E dA E r

Q Q Q QE r E N C

r

0 0

2

2

0 2

0

2

0

/ /

4

4 /4

4

encq Q

E dA E r

QE r Q E

r

Q qF Eq

r

0/ 0

0

0

encq

E dA

F Eq

(a) Derive the expression for the electric field magnitude in terms of the distance r from the center for the region r < a.

(b) Derive the expression for the electric field magnitude in terms of the distance r from the center for the region a < r < b. r is inside the conductor, thus E=0 (c) Derive the expression for the electric field magnitude in terms of the distance r from the center for the r > b.

(d) What is the surface charge density on the inner surface the shell?

24inner

inner

Q Q

A a

(e) What is the surface charge density on the outer surface the shell?

2

2

4outer

inner

Q Q

A b

0

0

2

2

2

0 0

/

4

44

enc

Qq

E dA E r

Q QE r E

r

0

0

2

2

2

0 0

3/

4

34

4

enc

Q Qq

E dA E r

Q Q QE r E

r

(4) A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d (see the Figure ). The inner shell has total charge +2q, and the outer shell has charge +4q. Express all answers in terms of q, r, a, b, c, d, π, and Є0. (a) Calculate the magnitude of the electric field in terms of q and distance r from the common center of the two shells for r < a. There is no enclosed charge, thus E=0 (b) Repeat part a with a < r < b. The Gaussian radius is inside the conductor, thus E=0 (c) Repeat part a with b < r < c.

00

2

22

0 0

2 4/

4

2 4 2 44

4

enc

q qq

E dA E r

q q q qE r E

r

(d) Repeat part a with c < r < d. The Gaussian radius is inside the conductor, thus E=0 (e) Repeat part a with r > d. (f) What is the total charge on the inner surface of the small shell? (g) What is the total charge on the outer surface of the small shell? (h) What is the total charge on the inner surface of the large shell? (i) What is the total charge on the outer surface of the large shell? (e)-(i):

(5) A very long uniform line of charge has charge per unit length λ1 = 4.72 µC/m and lies along the x-axis. A second long uniform line of charge has charge per unit length λ2 = -2.38 µC/m and is parallel to the x-axis at y1 = 0.402 m (a) What is the magnitude of the net electric field at point y2 = 0.202 m on the y-axis? What direction is the net electric field pointing along the y-axis?

Draw to Gaussian surfaces. Each Gaussian surface will help you find an E field. Add the two E fields to find the net electric field.

51 21 2

2 0 1 2 0

)2.06(10 /2 2 ( )netE E E N C

y y y

pointing in the +y direction

(b) What is the magnitude of the net electric field at point y3 = 0.602 m on the y-axis? What direction is the net electric field pointing along the y-axis?

Draw to Gaussian surfaces. Each Gaussian surface will help you find an E field. Add the two E fields to find the net electric field.

4

3

1 21 2

3 0 1 0

7.3(10 ) /2 2 ( )netE E E N C

y y y

pointing towards the –y

direction (6) Two very large, non-conducting plastic sheets, each 10.0 cm thick, carry uniform charge densities σ1, σ2, σ3, and σ4 on their surfaces, as shown in the figure . These

surface charge densities have the values σ1 = -6.00 µC/m, σ2 = +5.00 µC/m, σ3 = +2.00 µC/m, and σ4 = +4.00 µC/m. (a) Use Gauss's law to find the magnitude and direction of the electric field at the point A, 5.00 cm from the left face of the left-hand sheet.

Compress the left slab with the right slab to get a uniform charge density, add them and find the E field magnitude and direction

1 2 3 4( ) 5 /

netC m

5

0

2.82(10 ) /2

netE N C

pointing to the left

(b) Find the magnitude and direction of the electric field at the point B, 1.25 cm from the inner surface of the right-hand sheet.

Compress the left slab. Compress the right slab. Find the E field associated with each infinite sheet. Add the E fields.

1 1 21 /

netC m

2 43 ( 6 /)net

C m

51 2

0

3.95(10 ) /2

net netE N C

pointing to the left

(c) Find the magnitude of the electric field at the point C, in the middle of the right-hand sheet.

Compress all the infinite sheets to the left of point c into one infinite sheet. Add the electric fields due to the left and right sheets.

1 1 2 31 /

netC m

51

0

4 1.69(10 ) /2

( )netE N C

pointing to the left