from chapter 23 – coulomb’s law

From Chapter 23 – Coulomb’s Law

We found the fields in the vicinity of continuous charge distributions by integration:

2kE

x

Line of Charge:

o

E2

Charged Plate:

+

+

+

+

dE

x

rR

Are you interested in learning an easier way?

Electric flux

• The field is uniform• The plane is

perpendicular to the field

E E A

Flux for a uniform electric field passing through an arbitrary plane

• The field is uniform• The plane is not

perpendicular to the field

E E A E A cos

n̂

ˆA An

E E A

Rules for drawing field lines• The electric field, , is tangent to the field lines.• The number of lines leaving/entering a charge is

proportional to the charge.• The number of lines passing through a unit area normal

to the lines is proportional to the strength of the field in that region.

• Field lines must begin on positive charges (or from infinity) and end on negative charges (or at infinity). The test charge is positive by convention.

• No two field lines can cross.

E

# of electric field linesE

Area

EN E A

n̂

A

E

E

General flux definition

E ˆE n A E A cos

• The field is not uniform

• The surface is not perpendicular to the field

AIf the surface is made up of a mosaic of N little surfaces

N

E i i ii 1

ˆE n A

E ˆE ndA E dA

Electric Flux, General

• In the more general case, look at a small area element

• In general, this becomes

cosE i i i i iE A θ E A

0surface

limi

E i iAE A d

E A

Electric Flux Calculations

• 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/C2

Electric Flux, Closed Surface

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

are perpendicular to the surface

– By convention, they point outward

Closed surfaces

–+ Q – 3Q+ +Q

E ˆE ndA E dA

n̂ or A point in the direction outward from the closed surface

Negative FluxZero Flux

Flux from a point charge through a closed sphere

E ˆE ndA E dA

ˆ ˆE n E n E

2

kqˆE r E constant

r

E sphereˆE ndA E dA E A

2

E 2o

kq q4 r 4 kq

r

Gauss’s Law

Gauss asserts that the proceeding calculation for the flux from a point charge is true for any charge distribution!!!

enclosedE enclosed

o

qE dA 4 kq

This is true so long as Q is the charge enclosed by the surface of integration.

A few questions

• If the electric field is zero for all points on the surface, is the electric flux through the surface zero?

• If the electric flux is zero, must the electric field vanish for all points on the surface?

• If the electric flux is zero for a closed surface, can there be charges inside the surface?

• What is the flux through the surface shown? Why?

–+ Q – 3Q+enclosed

Eo

qE dA

Example P24.1

3 2cos 3.50 10 0.350 0.700 cos0 858 N m CE EA

An electric field with a magnitude of 3.50 kN/C is applied along the x axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long assuming that

(a) the plane is parallel to the yz plane; (b) the plane is parallel to the xy plane; (c) the plane contains the y axis, and its normal

makes an angle of 40.0° with the x axis.

90.0 0E

3 23.50 10 0.350 0.700 cos40.0 657 N m CE

Example P24.9

The following charges are located inside a submarine: 5.00 μC, –9.00 μC, 27.0 μC, and –84.0 μC.

(a) Calculate the net electric flux through the hull of the submarine.

(b) Is the number of electric field lines leaving the submarine greater than, equal to, or less than the number entering it?

6 2 2in12 2 2

0

5.00 C 9.00 C 27.0 C 84.0 C6.89 10 N m C

8.85 10 C N mE

q

Example P23.17A point charge Q = 5.00 μC is located at the center of a cube of edge L = 0.100 m. In addition, six other identical point charges having q = –1.00 μC are positioned symmetrically around Q as shown in Figure P24.17. Determine the electric flux through one face of the cube.

6 22

12 2one face0

6 5.00 6.00 10 C N m18.8 kN m C

6 6 8.85 10 CE

Q q

Applying Gauss’s Law

• Select a surface– Try to imagine a surface where the electric field is constant

everywhere. This is accomplished if the surface is equidistant from the charge.

– Try to find a surface such that the electric field and the normal to the surface are either perpendicular or parallel.

• Determine the charge inside the surface

• If necessary, break the integral up into pieces and sum the results.– e.g. For a cylinder,

2

top bottom side

dA dA dA dA 2 r 2 rh

r

h

L

r

Example – a line of charge

total charge

total length

E

1. Find the correct closed surface2. Find the charge inside

that closed surface

enclosedE

o

qE dA

Example – a charged plane

+Q

total charge

total area

1. Find the correct closed surface2. Find the charge inside

that closed surface

enclosedE

o

QE dA

E

r

Example – a solid sphere of charge

+Q uniformly distributed total charge

total volume

a

Inside the charged sphere:

r a

1. Find the correct closed surface2. Find the charge inside

that closed surface

enclosedE

o

qE dA

E

Example – a solid sphere of charge

+Q uniformly distributed total charge

total volume

a

E

Outside the charged sphere:

r a

1. Find the correct closed surface2. Find the charge inside

that closed surface

enclosedE

o

qE dA

r

Example – a conducting conductor

• Insulators, like the previous charged sphere, trap excess charge so it cannot move.

• Conductors have free electrons not bound to any atom. The electrons are free to move about within the material. If excess charge is placed on a conductor, the charge winds up on the surface of the conductor. Why?

• The electric field inside a conductor is always zero.

• The electric field just outside a conductor is perpendicular to the conductors surface and has a magnitude,

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

Einside = 0, cont.

• Before the external field is applied, free electrons are distributed throughout the conductor

• When the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external field

• There is a net field of zero inside the conductor• This redistribution takes about 10-15s and can be

considered instantaneous

Charge Resides on the Surface

• Choose a gaussian surface inside but close to the actual surface

• The electric field inside is zero (prop. 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

Charge Resides on the Surface, cont

• Since no net charge can be inside the surface, any net charge must reside on the surface

• Gauss’s law does not indicate the distribution of these charges, only that it must be on the surface of the conductor

Field’s Magnitude and Direction

• Choose a cylinder as the gaussian surface

• The field must be perpendicular to the surface– 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

Field’s Magnitude and Direction, cont.

• The net flux through the gaussian surface is through only the flat face outside the conductor– The field here is perpendicular to the

surface

• Applying Gauss’s law

Eo o

σA σEA and E

ε ε

Conductors in Equilibrium, example

• The field lines are perpendicular to both conductors

• There are no field lines inside the cylinder

r

Example – a solid conducting sphere of charge surrounded by a conducting shell

+2Q on inner sphere-Q on outer shell

a

Find the Electric Field:

r a

1. Find the correct closed surface2. Find the charge inside

that closed surface

enclosedE

o

qE dA

E

b

c

r ca r b

b r c

Example P24.31

Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field

(a) 10.0 cm and (b) 20.0 cm from the center of the

charge distribution.

9 6

2 2

8.99 10 32.0 107.19 MN C

0.200ekQEr

Example P24.35A uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 μC. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder.

9 2 2 62 8.99 10 N m C 2.00 10 C 7.00 m20.100 m

ekEr

51.4 kN C, radially outwardE

cos 2 cos0E EA E r 4 25.14 10 N C 2 0.100 m 0.0200 m 1.00 646 N m CE

Example P24.43

A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find

(a) the charge density of each face of the plate and

(b) the total charge on each face.

4 12 7 28.00 10 8.85 10 7.08 10 C m

277.08 10 0.500 CQ A 71.77 10 C 177 nCQ

Example P24.47A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of λ, and the cylinder has a net charge per unit length of 2λ. From this information, use Gauss’s law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder, a distance r from the axis.

0

2 3 6 3 radially outward

2e ek k

Er r r

in0 q inq

3