Unit 4, Slide 1

Physics 2112 Unit 4: Gauss’ Law

Today’s Concepts:A) ConductorsB) 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 Guass Law

Unit 4, Slide 3

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 4

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

Unit 4, Slide 5

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?

Unit 4, Slide 6

An infinite line of charge with linear density λ1 = -5.1μC/m is positioned along the axis of a thick conducting shell of inner radius a = 2.7 cm and outer radius b = 5.0 cm and infinite length. The conducting shell is uniformly charged with a linear charge density λ 2 = +2.0 μC/m.

Example 4.2

C) What is Ex(R), the electric field at point R, located a distance dR = 7.0 cm from the origin and making an angle of 30o with respect to the y-axis as shown?

x R

Unit 4, Slide 7

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

Unit 4, Slide 8

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 9

Gauss’ Law Symmetries

ALWAYS TRUE!

In cases with symmetry can pull E outside and get

Spherical Cylindrical Planar

0A

QE enc

24 rA

024 r

QE enc

rLA 2

02 r

E

22 rA

02

E

0

encQAdE

Electricity & Magnetism Lecture 4, Slide 10

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+1/2aB. a cube of dimension R+1/2a C. a cylinder with cross sectional radius of R+1/2a and arbitrary lengthD. This field cannot be calculated using Gauss' lawE. None of the above

Electricity & Magnetism Lecture 4, Slide 11

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

Electricity & Magnetism Lecture 4, Slide 12

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?

Quick Review: Infinite Sheet of Charge

Unit 4, Slide 13

Unit 4, Slide 14

CheckPoint: Infinite Sheets of Charge

In both cases shown below, the colored lines represent positive (blue) and negative (red) charged planes. The magnitudes of the charge per unit area on each plane is the same. In which case is the magnitude of the electric field at point P bigger?

A. Case AB. Case BC. They are the same

Electricity & Magnetism Lecture 4, Slide 15

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

x+3Q

Electricity & Magnetism Lecture 4, Slide 16

Calculation

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

What is E everywhere?

r < r1

r1 < r < r2

r > r2

20

3

4

1

r

QE

0E

neutral conductor

r1

r2

y

x+3Q

0

encQAdE