unit 4: gauss’ law today’s oncepts · gauss’ law always true! in cases with symmetry can pull...

Unit 4, Slide 1

Physics 2112

Unit 4: Gauss’ Law

Today’s Concepts:

A) Conductors

B) Using Gauss’ Law

Unit 4, Slide 2

Use #1: Determining E without calculus

E from infinite line of charge.

We did this before with calculus.

Remember?

22

2

0)tan*(

sin*sec*)(

hh

hdkPE

f

x

Let’s do it again using Gauss Law

o

liner

E

2

A solid conducting sphere

with net charge –Q and a

radius r1. What is the

electric field within the

sphere some distance r2

from the center? (r2<r1)

-Q

r1r2

A. -kQ/r12

B. -kQ/r22

C. -kQ(r2/r1)2

D. Zero

E. None of the above

Unit 3, Slide 3

Use #2: Determine Charge on Surfaces

Unit 4, Slide 4

E 0 inside any conductor at equilibrium

(If E ≠0, then charge feels force and moves!)

Excess charge on conductor only on surface at equilibrium

Why? Apply Gauss’ Law

E 0

Use #2: Determine Charge on Surfaces

o

enc

surface

QAdE

0

0encQ

Take Gaussian surface to be just inside conductor surface

Unit 4, Slide 5

ALWAYS TRUE!

How Does This Work?

Charges in conductor move to surfaces to make Qenclosed 0.

We say charge is induced on the surfaces of conductors

Gauss’ Law + Conductors + Induced Charges

o

enc

surface

QAdE

Question

Unit 4, Slide 6

An 2m x 3m sheet of plastic

has a surface charge

density s = 6uC/m2.

What is the linear charge density of the plate, λx, in the

x direction?

A) x = 36 μC/m

B) x = 18 μC/m

C) x = 12 μC/m

D) x = 6 μC/m

E) x = 0

2m

3mx

Question

Unit 4, Slide 7

An 3m x 3m sheet of plastic

has a linear charge density in

the x direction of x = 9mC/m.

What is the surface charge density, s, of the sheet?

A) s = 81 mC/m2

B) s = 27 μC/m2

C) s = 9 μC/m2

D) s = 3 μC/m2

E) s = 0

3m

3mx

Unit 4, Slide 8

An infinite line of charge with linear

density λ1 = -5.1μC/m is positioned

along the axis of a neutral conducting

shell of inner radius a = 2.7 cm and

outer radius b = 5.0 cm and infinite

length.

Example 4.1

A) What is λb, the linear charge density on the outer

surface of the conducting shell ?

B) What is Ex(R), the electric field at point R, located a

distance dR = 0.9 cm from the origin and making an angle

of 30o with respect to the y-axis as shown?

Question

Unit 4, Slide 9




cylinder of inner radius a = 2.7 cm and


length.

What is λa, the linear charge density on the inner surface of

the conducting shell ?

A) λa = -10.2μC/m

B) λa = -5.1μC/m

C) λa = 0

D) λa = +5.1μC/m

E) λa = +10.2μC/m

Question

Unit 4, Slide 10






length.

What is λb, the linear charge density on the outer surface of

the conducting shell ?

A) λb = -10.2μC/m

B) λb = -5.1μC/m

C) λb = 0

D) λb = +5.1μC/m

E) λb = +10.2μC/m

Question

Unit 4, Slide 11






length.

What is sa, the surface charge density on the inner surface

of the conducting shell ?

A) sa = +5.1μC/m2

B) sa = (+5.1*2*0.027)μC/m2

C) sa = (+5.1/(2*0.027))μC/m2

D) sa = 0

Question

Unit 4, Slide 12






length.

What is sb, the surface charge density on the outter


A) sb = -5.1μC/m2

B) sb = (-5.1/(2*0.050))μC/m2

C) sb = (-5.1/(2*0.027))μC/m2

D) sb = 0

Question

Unit 4, Slide 13






length.

A) λ1 only

B) λ1 and λa

C) λa only

D) λ1, λa and λb

Let’s say we wanted to find Ex(R), the electric field at point

R, located a distance dR = 0.9 cm from the origin and

making an angle of 30o with respect to the y-axis as shown.

This value would depend on:

Question

Unit 4, Slide 14






length.

A) λ1 only

B) λ1 and λa

C) λa only

D) λ1, λa and λb

Let’s say we wanted to find Ex(R), the electric field at point

R, located a distance dR = 8 cm from the origin and making

an angle of 30o with respect to the y-axis as shown. This

value would depend on:

Unit 4, Slide 15






length.

Example 4.1

A) What is λb, the linear charge density on the outer


B) What is Ex(R), the electric field at point R, located a



Unit 4, Slide 16



along the axis of a thick conducting



length. The conducting shell is

uniformly charged with a linear charge

density λ 2 = +2.0 μC/m.

Example 4.2(a)

C) What is Ex(R), the electric field at point R, located a



x R

Unit 4, Slide 17

A hollow insulating spherical shell has

a total charge of Q=-8uC and a

uniform charge distribution. The shell

has an inner radius of a = 2.7 cm and

outer radius b = 5.0 cm

Example 4.3

C) What is |E|, the electric field at point R, located a

distance dR = 4.0 cm from the center of the sphere?

x R

Unit 4, Slide 18

CheckPoint: Charged Conducting Sphere & Shell 1

A positively charged solid conducting sphere is contained within a negatively charged conducting spherical shell as shown. The magnitude of the total charge on each sphere is the same. Which of the following statements best describes the electric field in the region between the spheres?

A. The field points radially outwardB. The field points radially inwardC. The field is zero

Unit 4, Slide 19

CheckPoint: Slightly changed

A positively charged solid conducting sphere is contained within a positively charged conducting spherical shell as shown. The magnitude of the total charge on each sphere is the same. Which of the following statements best describes the electric field in the region between the spheres?

A. The field points radially outwardB. The field points radially inwardC. The field is zero

+q

Unit 4, Slide 21

Gauss’ Law Symmetries

ALWAYS TRUE!

In cases with symmetry can pull E outside and get

Spherical Cylindrical Planar

0A

QE enc

24 rA

0

24 r

QE enc

rLA 2

02

rE

22 rA

02

sE

0

encQAdE

Quick Review: Infinite Sheet of Charge

Unit 4, Slide 22

Unit 4, Slide 23

Gauss’ Law

ALWAYS TRUE!

In cases with symmetry can pull E outside and get

In General, integral to calculate flux is difficult…. and not useful!

To use Gauss’ Law to calculate E, need to choose surface carefully!

1) Want E to be constant and equal to value at location of interest

OR

2) Want E dot A 0 so doesn’t add to integral

0A

QE enc

0

encQAdE

Electricity & Magnetism Lecture 4, Slide 24

CheckPoint: Gaussian Surface Choice

You are told to use Gauss' Law to calculate the electric field at a distance R away from a charged cube of dimension a. Which of the following Gaussian surfaces is best suited for this purpose?

A. a sphere of radius R+1/2a

B. a cube of dimension R+1/2a

C. a cylinder with cross sectional radius of R+1/2a and arbitrary length

D. This field cannot be calculated using Gauss' law

E. None of the above


CheckPoint: Charged Sphericlal Shell

A charged spherical insulating shell has inner radius a and outer radius b. The charge density on the shell is ρ. What is the magnitude of the E-field at a distance r away from the center of the shell where r < a?

A. ρ/εo

B. zeroC. ρ(b3-a3)/(3εor2)D. none of the above


Example 4.3

A charged spherical insulating shell has inner radius a and outer radius b. The charge density on the shell is ρ.

What is the magnitude of the E-field at a distance r away from the center of the shell where a< r < b?


Example 4.4

Point charge 3Q at center of neutral conducting shell of inner radius r1 and outer radius r2.

What is E for the three regions of: r < r1

r1< r < r2

r2 < r ?

neutral conductor

r1

r2

y

x3Q


Calculation

Point charge 3Q at center of neutral conducting shell of inner radius r1 and outer radius r2.A) What is E everywhere?

We know: magnitude of E is fcn of rdirection of E is along

neutral conductor

r1

r2

y

x3Q r̂

r < r1

24 rEAdE

QQenc 3

r > r2

2

0

3

4

1

r

QE

2

20 )(

3

4

1

rr

QE

0E

A)

B)

C)

r1 < r < r2

2

0

3

4

1

r

QE

2

10

3

4

1

r

QE

0E

A)

B)

C)

We can use Gauss’ Law to determine EUse Gaussian surface = sphere centered on origin

0

encQAdE

2

0

3

4

1

r

QE


Calculation


What is E everywhere?

r < r1

r1 < r < r2

r > r2

2

0

3

4

1

r

QE

0E

neutral conductor

r1

r2

y

x3Q

0

encQAdE

Question



A)

B)

C)

D)

What can we say about the relative magnitudes of s1 and s2?

|||| 21 ss <

neutral conductor

r1

r2

y

x3Q

2

1

14

3

r

Q

s

0

encQAdE

|||| 21 ss

|||| 21 ss >

0|| 2 s

Question

Unit 4, Slide 31

Now suppose we give conductor a charge of Q.

What is the surface charges at r1?

r1

r2

+2Q

3Q

chargedconductor

r1

r2

y

x3Q

-Q

2

1

14

3

r

Q

s

2

1

14 r

Q

s

2

1

14

2

r

Q

s

A)

B)

C)

Question

Unit 4, Slide 32

Now suppose we give conductor a charge of Q.

What is the surface charges at r2?

r1

r2

+2Q

3Q

chargedconductor

r1

r2

y

x3Q

-Q

2

2

24

3

r

Q

s

2

2

24 r

Q

s

2

2

24

2

r

Q

s

A)

B)

C)

A positive charge is kept (fixed) at the center inside a fixed

spherical neutral conducting shell. Which of the following

represents the charge distribution on the inner and outer walls

of the shell?

A. B. C. D.

Question

The positive charge is now moved and kept fixed off-center

inside the fixed spherical neutral conducting shell. Which of the

following represents the charge distribution on the inner and

outer surfaces of the shell?

A. B. C.

D. E.

Question

The positive charge +Q is now kept fixed at the center of a

spherical neutral conducting shell. A negative charge –Q is

brought near the outside of the sphere. Which of the following

represents the charge distributions?

A. B.

C.D.

Question

unit 4: gauss’ law today’s oncepts · gauss’ law always true! in cases with symmetry can pull...

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