Hw: All Chapter 5 problems and exercises

Test 1 resultsAverage 75Median 78

>90 >80 >70 >60 >50 <50

521 – 79522 – 76523 – 73.3524 – 76.5525 – 76526 – 69.7

Outline

• Applications of Gauss’s Law- The single Fixed Charge- Field of a sphere of charge- Field of a spherical shell- A Line of Charge• Conductors and Insulators• The electric field of a conductor• The field in the cavity of a conductor;

Faraday’s Cage

Solid conducting sphere with charge Q

A

E

r

204

1

r

QEAr

0 EAr

V

r

A

A

Q

04

1

r

QVAr

04

1

A

QVAr

04

1

Electric field of a ball of charge

30

20

4

1

4

1

R

rQERr

r

QERr

Q

Electric field outside of a charged sphere is exactly the same as the electric field produced by a point charge, located at the center of the sphere, with charge equal to the total charge on the sphere.

Electric field of a spherical shell

Q

The field outside the shell is like that of a point charge, while the field everywhere inside the shell is zero.

Electric field of a line of charge

rE

02

A Charged, Thin Sheet of Insulating Material

++

+

+

++

+

++

++

02

E

Conductors and insulators

Charges reside at the surface of the conductor

Conductor

E=0+

+ ++

+

+++

++

+++

+

+

What have we learned about conductors?

• There is no electric field inside a conductor• Net charge can only reside on the surface

of a conductor • Any external electric field lines are

perpendicular to the surface (there is no component of electric field that is tangent to the surface).

• The electric potential within a conductor is constant

Electric field near a surface of a conductor

al

cap

EaEdSSdE

0a

Ea 0

E

d

+

+

+

+

+

+

a

l

-

-

-

cap

EaEdSSdE

0a

Ea 0

E (the total field at any point

between the plates)

Two parallel conducting plates

An Apparent Contradiction

+

+

+

+

+

+ -

-

-

0

E

An Apparent Contradiction

+

+

+

+

+

+

-

-

-

0

E

?2 0

E Near the surface of any conductor in electrostatics

0

E

2

2 02

E

1) There is a conducting spherical shell, inner radius A and outer radius B. If you put a charge Q on it, find the charge density everywhere.

2) There is a conducting spherical shell, inner radius A and outer radius B. A charge Q is put at the center. If you put a charge Q2 on the shell, find the charge density everywhere.

A sphere of radius A has a charge Q uniformly spread throughout its volume. Find the difference in the electric potential, in other words, the voltage difference, between the center and a point 2A from the center.

There is a conducting spherical shell, inner radius A and outer radius B. A charge Q1 is put at the center. If you now put charge -2Q1 on the shell, find the charge density at r=A and r=B.

2

1

120

r

r

rr rdEVV

since inside the conductor.0E

For any two points and inside the conductor 1r

2r

21 rr VV

The conductor’s surface is an equipotential.

Equipotential Surfaces

An equipotential surface is a surface on which the electric potential V is the same at every point.

Because potential energy does not change as a test charge moves over an equipotential surface, the electric field can do no work on such a charge. So, electric field must be perpendicular to the surface at every point so that the electric force is always perpendicular to the displacement of a charge moving on the surface.

Field lines and equipotential surfaces are always mutually perpendicular.

Eq

Method of images: What is a force on the point charge near a conducting plate?

-

-

-

-

-

-

-

-

-

-

-

-

-

Equipotential surface

The force acting on the positive charge is exactly the same as it would be with the negative image charge instead of the plate.

2

2

0 )2(4

1

a

qF

The point charge feels a force towards the plate with a magnitude:

a

Method of images: A point charge near a conducting plane.

?E

-

-

-

-

-

-

-

-

-

-

-

-

-

Equipotential surface

ar

P

23220 )(4

1

ra

aqE

23220 )(4

1

ra

aqE

23220 )(4

2

ra

aqE

Equilibrium in electrostatic field: Earnshaw’s theorem

There are NO points of stable equilibrium in any electrostatic field!

How to prove it? Gauss’s Law will help!

P

Imaginary surface surrounding P

If the equilibrium is to be a stable one, we require that if we move the charge away from P in any direction, there should be a restoring force directed opposite to the displacement. The electric field at all nearby points must be pointing inward – toward the point P. But that is in violation of Gauss’ law if there is no charge at P.

Thomson’s atom1899

If charges cannot be held stably, there cannot be matter made up of static point charges (electrons and protons) governed only by the laws of electrostatics. Such a static configuration would collapse!

CapacitorsConsider two large metal plates which are parallel to each other and separated by a distance small compared with their width.

Area A

The field between plates is 0

E

VL

y

LdydyEbottomVtopVLL

y00 00

)]()([

A

QLL

A

ALbottomVtopV

000

)]()([

0AQL

V

The capacitance is:

L

A

AQLQ

V

QC 0

0

Have a great day!