Physics 2 lecture 2 gauss s law

• In order to compute the electric flux, we divide the surface into a large number of infinitesimal area elements • The electric flux through an area element is • The total flux through

Gauss’s Law Lecture 2

 Gauss’s Law

 Conductors in Electrostatic Equilibrium

In a table-top plasma ball, the colorful lines emanating from the sphere give evidence of strong electric fields Using Gauss’s law, we show in this chapter that the electric field surrounding a charged sphere is identical to that of a point charge (Getty Images)

 A very useful computational technique

 Two of Maxwell’s Equations

 Gauss’s Law –The Equation

0

in surface

Closed

d



q A

of charge inside

0

in surface

Closed

d



q A

E

n A

A  



Air flow rate (flux) through A

vA A



Electric flux

n A

A  



• We showed that the strength of an electric field is proportional to the number of

field lines per area

• Electric flux Φ Eis proportional to the number of electric field lines penetrating

some surface.

EA EA

A E

0 cos

E

Electric flux

In general, a surface S can be curved and

the electric field E may vary over the

surface

• In order to compute the electric flux, we divide

the surface into a large number of infinitesimal

area elements

• The electric flux through an area element is

• The total flux through the entire surface can be

obtained by summing over all the area elements

6

• In order to compute the electric flux, we divide

the surface into a large number of infinitesimal

area elements

• The electric flux through an area element is

• The total flux through the entire surface can be

obtained by summing over all the area elements

i i

i i

Electric flux

A rectangle is an open surface —it does NOT contain a volume

A sphere is a closed surface —it DOES contain a volume

We are interested in closed surface

Gauss’s law

For a closed surface the normal vector is chosen

to point in the outward normal direction.

out points

Gauss’s law E 0 if E  points out







Gauss’s law

The net electric flux through any closed surface S is

where qin represents the net charge inside the surface

and E represents the electric field at any point on the

surface S.

0 surface

in S

E

q A

d



Gauss’s Law

Consider several closed surfaces surrounding a

charge q, as shown in figure above The number of

field lines through S 1 is equal to the number of field

lines through the surfaces S 2 and S 3 .



q A

d E A

d E A

d E

         

The number of electric field lines entering the surface equals the number leaving the surface The net flux is zero.

d

E  

Gauss’s Law

Gauss’s Law and Coulomb’s Law

 Around q we have drawn a spherical surface S of

radius r centered at q.

 At any point on the surface S, E is parallel to the

vector dA, thus

 Applying Gauss’s law, we have

 By symmetry, E is constant over the surface S, thus

 The electric field E at any point on surface S is of the

form

A E A

E A

d E

S S

 At any point on the surface S, E is parallel to the

vector dA, thus

 Applying Gauss’s law, we have

 By symmetry, E is constant over the surface S, thus

 The electric field E at any point on surface S is of the

E A

d E

S S

d



 r q E

A E

A E

S S





Applying Gauss’s Law

1 Identify the symmetry associated with the charge distribution.

2 Identify regions in which to calculate E field.

3 Determine the direction of the electric field, and a “Gaussian surface”

on which the magnitude of the electric field is constant over portions of the surface.

4 Calculate

5 Calculate q, charge enclosed by surface S.

6 Apply Gauss’s Law to calculate E:

1 Identify the symmetry associated with the charge distribution.

2 Identify regions in which to calculate E field.

3 Determine the direction of the electric field, and a “Gaussian surface”

on which the magnitude of the electric field is constant over portions of the surface.

4 Calculate

5 Calculate q, charge enclosed by surface S.

6 Apply Gauss’s Law to calculate E:

d E

   

Planar Symmetry

Consider an infinitely large non-conducting

plane with uniform surface charge density σ

Determine the electric field everywhere in space

1 An infinitely large plane possesses a planar symmetry

2 The electric field must point perpendicularly away from the

plane The magnitude of the electric field is constant on

planes parallel to the non-conducting plane

Planar Symmetry

(3) The Gaussian surface consists of three

parts: a two ends S1and S2plus the curved

side wall S3 The flux through the Gaussian

surface is

(4) The amount of charge enclosed by the

Gaussian surface is q= σA.

(5) Applying Gauss’s law gives

EA A

E A

E

A d E A

d E A

d E A

d E

E

S S

S S

E

20

2 2 1

1

3 3 2

2 1

1

3 2

(3) The Gaussian surface consists of three

parts: a two ends S1and S2plus the curved

side wall S3 The flux through the Gaussian

surface is

(4) The amount of charge enclosed by the

Gaussian surface is q= σA.

(5) Applying Gauss’s law gives

EA A

E A

E

A d E A

d E A

d E A

d E

E

S S

S S

E

20

2 2 1

1

3 3 2

2 1

1

3 2

Cylindrical Symmetry

An infinitely long rod of negligible radius has a

uniform positive charge density λ Find the electric

field a distance r from the rod.

(1) An infinitely long rod possesses cylindrical

symmetry

(2) The electric field must be point radially

away from the symmetry axis of the rod

The magnitude of the electric field is

constant on cylindrical surfaces of radius r.

Therefore, we choose a coaxial cylinder as

our Gaussian surface

(1) An infinitely long rod possesses cylindrical

symmetry

(2) The electric field must be point radially

away from the symmetry axis of the rod

The magnitude of the electric field is

constant on cylindrical surfaces of radius r.

Therefore, we choose a coaxial cylinder as

our Gaussian surface

Cylindrical Symmetry

(3) The Gaussian surface consists of three

parts: a two ends S1and S2plus the curved

side wall S3 The flux through the Gaussian

rl E

A E

A d E A

d E A

d E A

d E

s E

S S

S S

E



2d

0

3 3 2

2 1

1

3

3 2

(3) The Gaussian surface consists of three

parts: a two ends S1and S2plus the curved

side wall S3 The flux through the Gaussian

rl E

A E

A d E A

d E A

d E A

d E

s E

S S

S S

E



2d

0

3 3 2

2 1

1

3

3 2

y y

q

4 4

1

2 2

0 0

j l

q y

j l

y

q

0 0 0

0

2 0

1 4

Spherical Symmetry

A thin spherical shell of radius a has a charge +Q

evenly distributed over its surface Find the electric

field both inside and outside the shell.

• The charge distribution is spherically symmetric,

with a surface charge density

• The electric field must be radially symmetric and

• The charge distribution is spherically symmetric,

with a surface charge density

• The electric field must be radially symmetric and

Case 1: Inside the shell r < a.

• We choose our Gaussian surface to be a sphere of

q A

d

E 

a r

E  ,0 

Spherical Symmetry

Case 2: outside the shell r > a.

• We choose our Gaussian surface to be a sphere of

radius r, with r > a.

• Applying Gauss’s law

0 0

2 2

0

44

r E

q A

d E

E

in S

2 2

0

44

r E

q A

d E

E

in S

2 0

a E







Note that the field outside the sphere is the same as

if all the charges were concentrated at the center of

the sphere.

Spherical Symmetry

An insulating solid sphere of radius a has a uniform volume charge density ρ and carries

a total positive charge Q.

(a) Calculate the magnitude of the electric field at a point outside the sphere.

Solution

• The charge distribution is spherically symmetric, we

again select a spherical gaussian surface of radius r > a,

concentric with the sphere, as shown in Figure a

• Applying Gauss’s law

Solution

• The charge distribution is spherically symmetric, we

again select a spherical gaussian surface of radius r > a,

concentric with the sphere, as shown in Figure a

• Applying Gauss’s law

0 2

E

q A

d E

E

in S

4

1

r

Q E





Note that the field outside the sphere is the same as

if all the charges were concentrated at the center of

Point charge

Spherical Symmetry

An insulating solid sphere of radius a has a uniform volume charge density ρ and carries

a total positive charge Q.

(b) Find the magnitude of the electric field at a point inside the sphere.

Solution

• We select a spherical gaussian surface having radius r

< a, concentric with the insulating sphere

(Fig b).

• Applying Gauss’s law

23

Solution

• We select a spherical gaussian surface having radius r

< a, concentric with the insulating sphere

E

q A

d E

d

3 '

V

r V

V V





r a

Q r

a

Q r

3 0 0

Spherical Symmetry

An insulating solid sphere of radius a has a uniform volume charge density ρ and carries

a total positive charge Q.

The magnitude of the electric field at a point outside the sphere

The magnitude of the electric field at a point inside the sphere

2 0

4

1

r

Q E





r a

Conductors in Electrostatic Equilibrium

An insulator such as glass or paper is a material in which electrons are attached to some

particular atoms and cannot move freely.

 A good electrical conductor contains charges (electrons) that are not bound to any atom

and therefore are free to move about within the material.

When there is no net motion of charge within a conductor, the

conductor is in electrostatic equilibrium.

Conductors in Electrostatic Equilibrium

As we shall see, a conductor in electrostatic equilibrium has the following

properties:

1 The electric field is zero everywhere inside the conductor

2 If an isolated conductor carries a charge, the charge resides on its surface

3 The electric field just outside a charged conductor is perpendicular to the surface

of the conductor and has a magnitude σ/ε0 , where σ is the surface charge density at

that point

4 On an irregularly shaped conductor, the surface charge density is greatest at

locations where the radius of curvature of the surface is smallest

As we shall see, a conductor in electrostatic equilibrium has the following

properties:

1 The electric field is zero everywhere inside the conductor

2 If an isolated conductor carries a charge, the charge resides on its surface

3 The electric field just outside a charged conductor is perpendicular to the surface

of the conductor and has a magnitude σ/ε0 , where σ is the surface charge density at

that point

4 On an irregularly shaped conductor, the surface charge density is greatest at

locations where the radius of curvature of the surface is smallest

Conductors in Electrostatic Equilibrium

1 The electric field is zero everywhere inside the conductor

If the field was not zero, free charges in the conductor would accelerate under the action of the field Thus, the existence of electrostatic equilibrium is consistent only with a zero field in the conductor.

27

2 If an isolated conductor carries a charge, the charge resides on its surface.

The electric field is zero everywhere inside the conductor, thus

q A

d

E 

There can be no net charge inside the gaussian surface (which is

arbitrarily close to the conductor’s surface), any exess charge on

the conductor must reside on its surface.

Conductors in Electrostatic Equilibrium

3 The electric field just outside a charged conductor is perpendicular to the surface

of the conductor and has a magnitude σ/ε 0 , where σ is the surface charge density at

• Let’s consider a path integral of Eds around a closed path

shown in the figure

• If Et ≠ 0, the free charges would move along the surface; in

• The tangential component of E is zero on the surface of a

conductor.

• Let’s consider a path integral of Eds around a closed path

shown in the figure

• If Et ≠ 0, the free charges would move along the surface; in

0 d

d d

d



da cd

bc ab

s E s

E s

E s

E s

E          

0 0

d        ' 

 E  s  Et l En x En x

In the limit where both Δx and Δx’→ 0, we have E t Δl= 0.

Conductors in Electrostatic Equilibrium

3 The electric field just outside a charged conductor is perpendicular to the surface

of the conductor and has a magnitude σ/ε 0 , where σ is the surface charge density at

• The normal component of E has a magnitude σ/ε 0

• The tangential component of E is zero on the surface of a

E EA

q A

d E

n E

in S

That’s enough for today

 Please redo all the example problems given in your textbook

 Feel free to contact me via email

 Please redo all the example problems given in your textbook

 Feel free to contact me via email

See you all next week!