Electric flux Think of air blowing in through a window. How

much air comes through the window depends upon the speed of the air, the direction of the air, and the area of the window. We might call this air that comes through the window the "air flux".

Electric flux

The field is uniform

The plane is perpendicular to the field

E E A

We will define the electric flux for an electric field that is perpendicular to an area as

= 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 Acos

n

A AnE E A

n

A

E

E

General flux definition

E E n A E Acos

The field is not uniform

The surface is not perpendicular to the field

A If the surface is made up of a mosaic of N little surfaces

N

E i i i

i 1

E n A

E E 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 E A

0surface

limi

E i iA

E A d

E AIf we let the area of each element approach zero, then the number of elements approaches infinity and the sum is replaced by an integral.

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 Ai

point in different

directions

At each point, they

are perpendicular to

the surface

By convention, they

point outward

Closed surfaces

+ Q 3Q+ +Q

EE ndA E dA

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

Negative Flux(total charge negative) Zero Flux

Flux from a point charge through a closed sphere

EE ndA E dA

E n E n E

2

kqE r E constant

r

E sphereE ndA E dA E A

2

E 2

o

kq q4 r 4 kq

r

Gausss 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.

Example

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

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

(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 A point charge Q = 5.00 C 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 C are positioned symmetrically around Q as shown in Figure below. Determine the electric flux through one face of the cube.

0

6 22

12 2one face0

6

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

6 6 8.85 10 C

E total

E

Q q

Q q

Problem

An electric field is given by , where sign(x) equals 1 if x < 0, 0 if x = 0, and +1 if x > 0. A cylinder of length 20 cm and radius 4.0 cm has its center at the origin and its axis along the x axis such that one end is at x = +10 cm and the other is at x = 10 cm. (a) What is the electric flux through each end? (b) What is the electric flux through the curved surface of the cylinder? (c) What is the electric flux through the entire closed surface? (d) What is the net charge inside the cylinder?

( )(300 / )E sign x N C i

The field at both circular faces of the cylinder is parallel to the outward vector normal to the surface, so the flux is just EA. There is no flux through the curved surface because the normal to that surface is perpendicular to Er. The net flux through the closed surface is related to the net charge inside by Gausss law.

APPLICATION OF GAUSSS LAW TO CHARGED INSULATORS

Ways of choosing the gaussian surface

to determine the electric field

Ways of choosing the gaussian surface over which the surface integral given by Equation can be simplified and the electric field determined

EE ndA E dA

In choosing the surface, we should always take advantage of the symmetry of the charge distribution so that we can remove E from the integral and solve for it. The goal in this type of calculation is to determine a surface that satisfies one or more of the following conditions:

1. The value of the electric field can be argued by symmetry to be constant over the surface. 2. The dot product in Equation can be expressed as a simple algebraic product E dA because E and dA are parallel. 3. The dot product in Equation is zero because E and dA are perpendicular.

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.

L

r

Example a line of charge

total charge

total length

E

1. Find the correct closed surface 2. Find the charge inside that closed surface

enclosedE

o

qE dA

0

0

(2 )

2

insideqEA

LE rL

kE

r

Example

A uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 C. 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.

(2 )

charge over length of cylinder=7

(2 )7

27

4

enclosedE

o

E cylinder

enclosedr

o

enclosed

enclosedr

o o

r

o

qE dA

E ndA E dA E A

qE rL

Qq xL

q Q LE rL

Q

kE

r r

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

ekE

r

51.4 kN C , radially outw ardE

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 a charged plane

+Q

total charge

total area

1. Find the correct closed surface 2. Find the charge inside that closed surface

enclosedE

o

QE dA

E

Example a charged plane

+Q

total charge

total area

1. Find the correct closed surface(cylinder) 2. Find the charge inside that closed surface

enclosedE

o

QE dA

E0

0

0

2

2

2

insideqEA

AEA

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 surface 2. 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

EOutside the charged sphere:

r a

1. Find the correct closed surface 2. Find the charge inside that closed surface

enclosedE

o

qE dA

r

A non-conducting sphere of radius 6.00 cm has a uniform volume charge density of 450 nC/m3. (a) What is the total charge on the sphere? Find the electric field at the following distances from the spheres center: (b) 2.00 cm, (c) 5.90 cm, (d) 6.10 cm, and (e) 10.0 cm. Ans: (a)0.407 nc (b)339 N/C1KN/C(d)983N/C(e)366 N/C (e)

PROBLEM

A sphere of radius R has volume charge density = B/r for r < R , where B is a constant and = 0 for r > R. (a) Find the total charge on the sphere. (b) Find the expressions for the electric field inside and

outside the charge distribution (c) Sketch the magnitude of the electric field as a function

of the distance r from the spheres center.

PROBLEM

CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM

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, /o

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

Gausss law does not indicate the

distribution of these charges, only that it

must be on the surface of the conductor

Fields Magnitude and Direction

Choose a cylinder as the gaussian surface

The field must be perpendicular to the surface

Fields 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 Gausss law

E

o o

A EA and E

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 surface 2. Find the charge inside that closed surface

enclosedE

o

qE dA

E

b

c

r c

a r b

b r c

At inner surface of shell, gaussian surface Since E=0, Qenc=0=charge on inner sphere+charge on inner surface of shell Therefore charge on inner shell=-2Q

Example

Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 C 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 M N C

0.200

ekQE

r

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

A 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 , and the cylinder has a net charge per unit length of 2. From this information, use Gausss 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 outw ard

2

e ek k

Er r r

in0 q inq 3

Recall

Del

Grad

Vector operator acts on a scalar field to generate a vector field

Example

Div

Vector operator acts on a vector field to generate a scalar field

Example

Curl

Vector operator acts on a vector field to generate a vector field

Example

Gausss Theorem (differential form)

Applying the Divergence Theorem

The integral and differential form of Gausss Theorem

Qin

Gauss' divergence theorem relates triple integrals and surface integrals.

del operator

x y zx y z

: sin cosd d

x xdx dx

Examples of derivative operators:

scalar

vector

This is a vector operator.

: sin cos

sin sin cos

sin sin cos

d dx x x x x

dx dx

d dx x x x x x x

dx dx

d dx y x x y x z x

dx dx

Example

, , sin 3E x y z x x y y z xy

, , sin 3E x y z x y z x x y y z xyx y z

, , sin 3E x y z x y xyx y z

, , cos 3 0 3 cosE x y z x x

, ,E x y zFind

Example: Problem If the electric field in some region is given (in spherical coordinates) by the expression

where A and B are constants, what is the charge density ?