EM Assignment 1

Due: 2:00pm on Saturday, September 21, 2013

You will receive no credit for items you complete after the assignment is due. Grading Policy

Problem 20.64

An electron is at the origin, and an ion with charge 5 is at = 10 .

Part A

Find a point where the electric field is zero.

Express your answer using two significant figures.

ANSWER:

Problem 20.69

A straight wire 10 long carries 27 distributed uniformly over its length.

Part A

What is the line charge density on the wire?

Express your answer using two significant figures.

ANSWER:

Part B

Find the electric field strength 10 from the wire axis, not near either end.

Express your answer using two significant figures.

ANSWER:

Part C

Find the electric field strength 470 from the wire.

Note: make suitable approximation. Express your answer using two significant figures.

ANSWER:

+ e x nm

= x nm

m μC

= λ μC/m

cm

= E N/C

m

= E N/C

Problem 20.77

A thin rod extends along the axis from to , and carries a line charge density ,

where is a constant.

Part A

Find the electric field at the position .

Express your answer in terms of the variables , , , , and .

ANSWER:

Problem 20.78

A rod of length 2 lies on the axis, centered at the origin, and carries a line charge density given by , where is a constant.

Part A

Find an expression for the electric field strength at points on the axis, for .

Express your answer in terms of , , , , , .

ANSWER:

Part B

Show that for , your result has the dependence of a dipole field, and by comparison, determine

the dipole moment of the rod.

ANSWER:

Problem 21.69

A solid sphere of radius carries a volume charge density , where is a constant and is the

distance from the center.

x x = 0 x = L λ = λ0(x/L)2

λ0

x = − L

i j λ0 L k

= E

L xλ = (x/L)λ0 λ0

x x > L

i j λ0 L x k

= E

x ≫ L 1/x3

3785 Character(s) remaining

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

(none provided)

R ρ = ρ0 er/R ρ0 r

Part A

Find an expression for the electric field strength at the sphere's surface.

Express your answer in terms of , and .

ANSWER:

A Conducting Shell around a Conducting Rod

An infinitely long conducting cylindrical rod with a positivecharge per unit length is surrounded by a conductingcylindrical shell (which is also infinitely long) with a chargeper unit length of and radius , as shown in thefigure.

Part A

What is , the radial component of the electric field between the rod and cylindrical shell as a function of the

distance from the axis of the cylindrical rod?

Express your answer in terms of , , and , the permittivity of free space.

You did not open hints for this part.

ANSWER:

Part B

What is , the surface charge density (charge per unit area) on the inner surface of the conducting shell?

You did not open hints for this part.

ANSWER:

ρ0 ϵ0 R

= E

λ

−2λ r1

E(r)r

λ r ϵ0

= E(r)

σinner

Part C

What is , the surface charge density on the outside of the conducting shell? (Recall from the problem

statement that the conducting shell has a total charge per unit length given by .)

You did not open hints for this part.

ANSWER:

Part D

What is the radial component of the electric field, , outside the shell?

You did not open hints for this part.

ANSWER:

A Charged Sphere with a Cavity

An insulating sphere of radius , centered at the origin, has a uniform volume charge density .

Part A

Find the electric field inside the sphere (for < ) in terms of the position vector .

Express your answer in terms of , (Greek letter rho), and .

Hint 1. How to approach the problem

Apply Guass's law, which states that for a closed surface, the integral of the scalar product of the

surface area and the electric field is directly proportional to the enclosed charge: .

Because this problem deals with a uniform spherically symmetric volume charge density, a logical

Gaussian surface in this case is a sphere of radius , with < . Find the integral (which will

involve ) and the enclosed charge. Then solve for .

Hint 2. Determine the enclosed charge

= σinner

σouter

−2λ

= σouter

E(r)

= E(r)

a ρ

( )E r r a r

r ρ ϵ0

∮ ⋅ d =E A qenclϵ0

r r a ∮ ⋅ dE A

( )E r E

What is the charge enclosed by a Gaussian sphere centered at the origin with radius (for < )?

Express your answer in terms of (the magnitude of ) and (Greek letter rho).

ANSWER:

Hint 3. Calculate the integral over the Gaussian surface

Because of the symmetry of the problem, the value of is constant over the entire Gaussian

surface. In particular, , where is the magnitude of the electric field at radius

, and is the surface area of the Gaussian sphere of radius . Find an expression for .

Express your answer in terms of and .

ANSWER:

ANSWER:

Part B

A spherical cavity is excised from the inside of thesphere. The cavity has radius and is centered at

position , where , so that the entire

cavity is contained within the larger sphere. Find theelectric field inside the cavity.

Express your answer as a vector in terms of any

or all of (Greek letter rho), , , and .

Hint 1. How to approach the problem

Use the principle of superposition. A region of zero charge behaves just like a region with equal amountsof positive and negative charge. Consider the field produced by an imaginary sphere the size of thecavity, with charge density opposite that of the larger sphere. If you add the field from the imaginarysphere to the field produced by the original, intact, sphere, you will obtain the field produced by thesphere with the cavity.

qencl r r a

r r ρ

= qencl

⋅ dE A

∮ ⋅ d = E(r)A(r)E A E(r)

r A(r) r A(r)

r π

= A(r)

= ( )E r

a4

h | | < ah 34

ρ ϵ0 r h

Hint 2. Find the field due to the imaginary sphere

Consider an imaginary sphere with charge density and radius centered at . Ignoring the actual

sphere, what is the field inside the imaginary sphere?

Express your answer as a vector in terms of , , (Greek letter rho) and .

Hint 1. Use the result for a uniformly charged sphere

The imaginary sphere is just like the uniformly charged sphere studied in Part A of this problem,except that it has a different charge density and different position. Therefore, you can use theresult you already obtained from the uniformly charged sphere if you use the new charge densityand if you replace with a new vector that represents the displacement from the center of the

imaginary sphere to .

What is in terms of and ?

ANSWER:

ANSWER:

ANSWER:

Score Summary:

Your score on this assignment is 0.0%.You received 0 out of a possible total of 7 points.

−ρ a4 h

( )E imag r

r h ρ ϵ0

r s r

s r h

= s

= ( )E imag r

= ( )E r