Spring 2015 Lecture 3 17.4-17 - Purdue University

PHYS 221 General Physics II

Spring 2015 Assigned Reading:

Lecture

Gauss’ Law, electric flux, conductors

17.4-17.53

REVIEW: Electric Field

Phys 221 S2015 Lecture 03 2

The electric field is a vector field: consists of a distribution of vectors, one for each point in the region around a charged object.

Test charge qo at point P

+

F

Electric field at point P

+

E

Define: Magnitude:

Direction:

E Fq0

Direction of that acts on a positive test charge

F

SI Unit: N/C

REVIEW: Electric Field Lines

• A visualization tool to illustrate the geometry of an electric field.


• Electric field lines originate from positive charges and terminates at negative charges.

• The direction of the electric field at any location is tangential to the field line there.

• The magnitude of the electric field at any location is proportional to the density of the lines there.

i>Clicker questionBB BB

A B

X

Y

The magnitude of the electric field at point X is greater than at point Y

A) True B) False (C) Don’t have cluePhys 221 S2015 Lecture 03 4

E

Fq0

F k

q0q1

r2r̂

E k

q1

r2r̂

E

E1 ...

En k

q1

r1r̂1 ... k

qn

rn

r̂n

• Last lecture: E field

• For a point like charge

• For a collection of charges

Today we look at another way to determine E Phys 221 S2015 Lecture 03 5

Gauss’ Law


… relates the electric fields at points on a closed Gaussian surface to the net charge enclosed by that surface.

What are in the boxes?

Concept of Flux


v

Air stream with uniform velocity, v, flows through a loop of area A

Av A xt volume

time

can define a flux:

• Flux is the amount of “something” (electric field, bananas, whatever you want) passing through a surface.

• The total flux depends on strength of the field, the size of the surface it passes through, and their orientation.

Flux of an Electric Field


arealines ofnumber ~E EA~lines ofnumber

area

Works only when lines are perpendicular to surface!

Need to quantify the pattern.More lines comes from the box – more charge inside!

Q: What is the number of lines that cross the surface?

Electric field is proportional to the number of lines per unit cross-sectional area:

Flux of an Electric Field


Analogy: think of rain which falls into a bucket.Less rainwater enters the bucket if it is tilted

AEEAEA cos~lines ofnumber

The number of lines crossing a surface is proportional to the perpendicular component of the field.

Definition of electric field flux:

cosEAAEEAE


Electric field flux

Field lines can either enter the closed surface area, or exit it.

•Flux is positive when lines leave the area

Flux sign is consistent with the sign of the enclosed net charge.

•Flux is negative when lines enter the area

Definition of electric field flux:

cosEAAEEAE

Example: Flux through a Cube


y

E

x

z

1 cos 0E EA

1

2 cos 0E EA 2

33 cos 0E EA

4 4 cos 0E EA 5

5 cosE EA EA 6 6 cosE EA EA

1 2 3 4 5 6 0E E E E E E EA EA

Consider a uniform electric field oriented parallel to the x -direction. Find the net flux through the surface of a cube of edge L oriented as shown.

No charge inside E =0

i>Clicker questionBB BB

• Consider two Gaussianspheres (of radius R and 2R )drawn around a single chargeas shown.

• Which of the followingstatements about the netelectric flux through the twosurfaces (2R and R) is true?

(a) R < 2R (b) R = 2R (c) R > 2R


Gauss’ Law


Relates net flux, , of an electric field through a closed surface to the net charge that is enclosed by the surface.

E EA EA qenc

o

Number of lines which exit (or enter) the closed area is proportional to the net charge inside:

Using Gauss’s law:• can determine charge inside an area from electric field on surface;• sometimes can find the field on surface based on charges inside

Gaussian Surface


• Gauss' law applies to any surface. • Construct an imaginary surface (Gaussian surface) to take

advantage of the symmetry and evaluate easily the electric field

Cylindrical geometry

Planar geometrySpherical geometry

E

Point-like charge


24E EA E r

20 0

2 20

4

4

E E r qq qE k

r r

Gauss’s Law and Coulomb’s Law are equivalent

+

Gaussian Surface

Gauss Law

-

+

Gauss’ Law


E EA EA qenc

o

4kqenc

-

• Gaus’ Law is not useful to determine the E field of a dipole.

• It tells us the flux is zero

Gauss’ Law: Spherical Symmetry


• A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at the center of the shell

Symmetry: 1. field should be radial2. The same at every location

on the spherical surface

0E

q

Gauss’s law:

24eo

qEA E r

24 o

qEr

r



Prove

•A shell of uniform charge exerts no electrostatic force on a charged particle that is located inside the shell.

0eo

qEA

rChoose Gaussian surface inside shell



Electric Field inside and outside a shell of uniform charge distribution

Demo: 5A-13 No Internal Field

Conductors versus Insulators


• Insulators: material in which electric charges are “frozen” in place.

• Glass, rubber, ceramic, quartz, plastic etc

• Conductors: material in which electric charges can move around “freely.

• Silver, copper, gold, aluminum, iron, etc

Conductors


• If an excess charge is placed on an isolated conductor, that amount of charge will move entirely to the surface of the conductor. None of the excess charge will be found within the body of the conductor.

E inside =0 otherwise we will have a current Therefore charge must be on the surface of the conductor

Conductors: Shielding from an Electrical Field


• Electric field lines for an oppositely charged metal cylinder and metal plate.

Note that:1. Electric field lines are

perpendicular to the conductors.

2. There are no electric field lines inside the cylinder.

3. The density of field lines is greatest at regions of greatest curvature

Conductors: Shielding from an Electrical Field


studentsensor

screened cage

sparks

Van de Graaffgenerator

Infinite Line of charge density


• From Symmetry: E-field only depends on distance r from line

• Select the Gaussian surface to be a cylinder of radius r and length haligned with the x-axis.

+ + + + ++ + + +x

y

+ + + + + + + + + + +

Er

h

Er

+ + + + + ++ + +

•Apply Gauss’ Law and assume uniform charge density :• On the ends,

and• On the barrel,

E EAcos 0

E EAcos 2rhE

E

20r

qenclosed h

0 E enclosedq

Spring 2015 Lecture 3 17.4-17 - Purdue University

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Transcript of Spring 2015 Lecture 3 17.4-17 - Purdue University