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Chapter 24

Gauss’s Law

Electric Flux

Electric flux is the

product of the

magnitude of the

electric field and the

surface area, A,

perpendicular to the

field

ΦE = EA

Defining Electric Flux

EFM06AN1

Electric Flux, General Area

The electric flux is

proportional to the

number of electric field

lines penetrating some

surface

The field lines may

make some angle θ

with the perpendicular

to the surface

Then ΦE = EA cos θ

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Electric Flux, Interpreting the

Equation

The flux is a maximum when the

surface is perpendicular to the field

The flux is zero when the surface is

parallel to the field

If the field varies over the surface, then

Φ = EA cos θ is valid for only a small

element of the area

Electric Flux, General

In the more general

case, look at a small

area element

In general, this

becomes

0

surface

limi

E i iA

E A d

E A

E E

iA

icos

i E

i A

i

Electric Flux, final

The surface integral means the integral must be evaluated over the surface in question

In general, the value of the flux will depend both on the field pattern and on the surface

The units of electric flux will be N.m2/C

i.e. [N/C] x [m2]

Electric Flux, Closed Surface

Assume a closed surface

The vectors ΔAi point in different directions At each point, they

are perpendicular to the surface

By convention, they point outward

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Flux Through Closed Surface

The net flux through the surface is

proportional to the net number of lines leaving

the surface

This is the number of lines leaving the surface

minus the number entering the surface

If En is the component of E perpendicular to

the surface, then

E E dA E

ndA

Quick Quiz 24.1

Suppose the radius of a sphere, with a charge at its centre, is

halved. What happens to the flux through the sphere and the

magnitude of the electric field at its surface?

(a) The flux and field both increase.

(b) The flux and field both decrease.

(c) The flux increases and the field decreases.

(d) The flux decreases and the field increases.

(e) The flux remains the same and the field increases.

(f) The flux decreases and the field remains the same.

Answer: (e). The same number of field lines pass through a

sphere of any size. Because points on the surface of the

sphere are closer to the charge, the field is stronger.

Quick Quiz 24.1 Quick Quiz 24.2

In a charge-free region of space, a closed container is placed

in an electric field. A requirement for the total electric flux

through the surface of the container to be zero is that

(a) the field must be uniform

(b) the container must be symmetric

(c) the container must be oriented in a certain way

(d) The requirement does not exist – the total electric flux is

zero no matter what.

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Answer: (d). All field lines that enter the container also leave

the container so that the total flux is zero, regardless of the

nature of the field or the container.

Quick Quiz 24.2

Gauss’s Law, Introduction

Gauss’s law is an expression of the

general relationship between the net

electric flux through a closed surface

and the charge enclosed by the surface

The closed surface is often called a

Gaussian surface

Gauss’s law is of fundamental

importance in the study of electric fields

Gauss’s Law – General

A positive point charge, q, is located at the centre of a sphere of radius r

The magnitude of the electric field everywhere on the surface of the sphere is E = keq / r2

Gauss’s Law – General, cont.

The field lines are directed radially outward

and are perpendicular to the surface at

every point

This will be the net flux through the

gaussian surface, the sphere of radius r

We know E = keq/r2 and Asphere = 4πr2,

E E A

sphere 4k

eq

q

0

E E dA E dA

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Gauss’s Law – General, notes

The net flux through any closed surface surrounding a point charge, q, is given by q/εo and is independent of the shape of that surface.

The net electric flux through a closed surface that surrounds no charge is zero.

Since the electric field due to many charges is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as

E E dA E

1 E

2 dA

Gauss’s Law – Final

Gauss’s law states

qin is the net charge inside the surface

E represents the electric field at any point on

the surface

E is the total electric field and may have contributions

from charges both inside and outside of the surface

Although Gauss’s law can, in theory, be solved

to find E for any charge configuration, in

practice it is limited to symmetric situations

E E dA

qin

0

Finding an electric field using Gauss’s law

EFA06AN1

Quick Quiz 24.3

If the net flux through a gaussian surface is zero, the

following four statements could be true. Which of the

statements must be true?

(a) There are no charges inside the surface.

(b) The net charge inside the surface is zero.

(c) The electric field is zero everywhere on the surface.

(d) The number of electric field lines entering the surface

equals the number leaving the surface.

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Answer: (b) and (d).

Statement (a) is not necessarily true because an equal

number of positive and negative charges could be present

inside the surface.

Statement (c) is not necessarily true - the number of field

lines entering the surface will be equal to the number

leaving the surface, but this does not mean the field is zero

on the surface.

Quick Quiz 24.3 Quick Quiz 24.4

Consider the charge distribution shown in the figure. The

charges contributing to the total electric flux through surface

S’ are

(a) q1 only

(b) q4 only

(c) q2 and q3

(d) all four charges

(e) none of the charges

Answer: (c). The charges q1 and q4 are outside the surface

and contribute zero net flux through S’.

Quick Quiz 24.4 Quick Quiz 24.5

Again consider the charge distribution shown in this figure.

The charges contributing to the total electric field at a chosen

point on the surface S’ are

(a) q1 only

(b) q4 only

(c) q2 and q3

(d) all four charges

(e) none of the charges

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Answer: (d). We don't need the surfaces to realize that any

given point in space will experience an electric field due to

all local source charges.

Quick Quiz 24.5 Conditions for a Gaussian

Surface

Choose a surface over which the integral can be simplified and the electric field determined.

Try to choose a surface that satisfies one or more of these conditions: The value of the electric field can be argued from

symmetry to be constant over the surface

The dot product of E.dA can be expressed as a simple algebraic product EdA because E and dA are parallel

The dot product is 0 because E and dA are perpendicular

The field can be argued to be zero over the surface

Field Due to a Point Charge

Choose a sphere as the

Gaussian surface

E is parallel to dA at each

point on the surface

E E dA E dA

q

0

E E dA E (4r

2)

E q

40r

2 k

e

q

r2

Field Due to a Spherically

Symmetric Charge Distribution

Select a sphere as the

Gaussian surface

For r >a

E E dA E dA

Q

0

E 4r2Q

0

so E Q

40r

2 k

e

Q

r2

Q is the total charge inside the surface

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Spherically Symmetric, cont.

Select a sphere as the Gaussian

surface, r < a

qenclosed = (4/3πr3)

Charge density

= Q / [4/3πa3]

E E dA E dA

qenc

0

E qenc

40r

2 k

e

qenc

r2 k

e

4

3r

3

r2

4k

er

3keQr

a3

Spherically Symmetric

Distribution, final

Inside the sphere, E

varies linearly with r

E → 0 as r → 0

The field outside the

sphere is equivalent

to that of a point

charge located at

the centre of the

sphere

Field Due to a Thin Spherical

Shell

Use spheres as the Gaussian surfaces

When r > a, the charge inside the surface is Q and

E = keQ / r2

When r < a, the charge inside the surface is 0 and E = 0

Field at a Distance from a Line

of Charge

Select a cylindrical charge distribution The cylinder has a

radius of r and a length of ℓ

E is constant in magnitude and perpendicular to the surface at every point on the curved part of the surface

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Field Due to a Line of Charge

The end view

confirms the field is

perpendicular to the

curved surface

The field through the

ends of the cylinder

is 0 since the field is

parallel to these

surfaces

Field Due to a Line of Charge

Use Gauss’s law to find the field

E E dA E dA EA

qencl

0

l

0

A 2rl, so E (2rl ) l

0

E

20r 2k

e

r

E must be perpendicular to the plane

and must have the same magnitude at

all points equidistant from the plane

Choose a small cylinder whose axis is

perpendicular to the plane for the

Gaussian surface

E is parallel to the curved surface, with

no contribution to the surface area from

this curved part of the cylinder

The flux through each end of the

cylinder is EA and so the total flux is

2EA

Field Due to a Plane of Charge Field Due to a Plane of Charge

The total charge in the surface is σA

Applying Gauss’s law

Note, this does not depend on r

Therefore, the field is uniform everywhere

E 2EA

A

0

, so E

20

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Electrostatic Equilibrium

When there is no net motion of charge

within a conductor, the conductor is said

to be in electrostatic equilibrium

Properties of a Conductor in

Electrostatic Equilibrium

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 and has a magnitude of σ/εo.

4. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest.

Property 1: Einside = 0

Consider a conducting slab in

an external field E

If the field inside the conductor

were not zero, free electrons in

the conductor would

experience an electrical force

These electrons would

accelerate

These electrons would not be

in equilibrium

Therefore, there cannot be a

field inside the conductor

Property 2: Charge Resides

on the Surface

Choose a Gaussian surface inside but close to the actual surface

The electric field inside is zero (property 1)

There is no net flux through the Gaussian surface

Because the Gaussian surface can be as close to the actual surface as desired, there can be no charge inside the surface

Therefore, any charge must be on the surface.

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Demo EA10: Gauss’s law; field inside insulator, not conductor

Excess charges resides on the outer surface of a conductor. A metal can is placed on top of a van der Graaff generator and filled with styrofoam packing. When the generator is turned on, the packing remains within the cup. One can show that the outside of the can has charge by drawing off a spark with the earth rod. The metal can and the generator are all at one equipotential and the charge is on the outside of the conductor. There is no field inside the can. The can is replaced by a polystyrene cup. In this case, the packing is ejected out of the cup. Because the cup is an insulator, a field can exist within it – the cup makes little disturbance to the equipotential surfaces.

Property 3: Field’s Magnitude

and Direction

Choose a cylinder as the Gaussian surface

The field must be perpendicular to the surface, for, if there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium.

E EA

A

0

, and E

0

Property 4: Charge density

highest at sharp points

The charge density is high

where the radius of curvature is

small

And low where the radius of

curvature is large

The electric field is large near

the convex points having small

radii of curvature and reaches

very high values at sharp points

End of Chapter

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Quick Quiz 24.6

Your little brother likes to rub his feet on the carpet and then

touch you to give you a shock. While you are trying to

escape the shock treatment, you discover a hollow metal

cylinder in your basement, large enough to climb inside. In

which of the following cases will you not be shocked?

(a) You climb inside the cylinder, making contact with the

inner surface, and your charged brother touches the outer metal

surface.

(b) Your charged brother is inside touching the inner metal

surface and you are outside, touching the outer metal surface.

(c) Both of you are outside the cylinder, touching its outer

metal surface but not touching each other directly.

Answer: (a). Charges added to the metal cylinder by your

brother will reside on the outer surface of the conducting

cylinder. If you are on the inside, these charges cannot

transfer to you from the inner surface. For this same reason,

you are safe in a metal automobile during a lightning storm.

Quick Quiz 24.6