The problem solving session will be Wednesdays from 12:30 – 2:30 (or until there is nobody left asking questions) in FN 2.212

Torque on the electric dipole

Electric field is uniform in space Net Force is zero Net Torque is not zero

( )( sin )qE d

p E

(torque is a vector)

Stable and unstable equilibrium

p E

p E

(electric dipole moment from “-” to “+”)

€

rp = q

r d

Charge #1

Charge #2

Charge #3

+q

+q

–q

x

y

A. clockwise.

B. counterclockwise.

C. zero.

D. not enough information given to decide

Three point charges lie at the vertices of an equilateral triangle as shown. Charges #2 and #3 make up an electric dipole.

The net electric torque that Charge #1 exerts on the dipole is

Electric field of a dipole

+-d

A E

1r

2r

E-field on the line connecting two charges

3

2 ep kE

r

when r>>d

E-field on the line perpendicular to the dipole’s axis

2E

+-d

AE

1E

General case – combination of the above two5 3

3( )p r pE r

r r

€

E = keq1

r22

−1

r12

⎛

⎝ ⎜

⎞

⎠ ⎟

€

E = 2E2 sinα

2= E2

d

r

E2 = ke

q

r2

E = ke

qd

r3

r E = −ke

r p

r3

Dipole’s Potential EnergyE-field does work on the dipole – changes its potential energyWork done by the field (remember your mechanics class?)

sindW d pE d

U p E

Dipole aligns itself to minimize its potential energy in the external E-field.Net force is not necessarily zero in the non-uniform electric field – induced polarization and electrostatic forces on the uncharged bodies

Chapter 22Gauss’s Law

Charge and Electric Flux

Previously, we answered the question – how do we find E-field at any point in space if we know charge distribution?

Now we will answer the opposite question – if we know E-fielddistribution in space, what can we say about charge distribution?

Electric flux

Electric flux is associated with the flow of electric field through a surface

For an enclosed charge, there is a connection between the amount of charge and electric field flux.

2

2

1~

~

Er

S r

E S const

Calculating Electric Flux

dVA

dt

Amount of fluid passing through the rectangle of area A

cosdV

Adt

dVA

dt

Flux of a Uniform Electric Field

cosE E A EA

A A n n

- unit vector in the direction of normal to the surface

Flux of a Non-Uniform Electric Field

ES

E d A

E – non-uniform andA- not flat

Few examples on calculating the electric flux

32 10 [ / ]E N C

Find electric flux

Gauss’s Law

0

i

E

q

E d A

Applications of the Gauss’s Law

If no charge is enclosed within Gaussian surface – flux is zero!

Electric flux is proportional to the algebraic number of lines leavingthe surface, outgoing lines have positive sign, incoming - negative

Remember – electric field lines must start and must end on charges!

Examples of certain field configurations

Remember, Gauss’s law is equivalent to Coulomb’s law

However, you can employ it for certain symmetries to solve the reverse problem – find charge configuration from known E-field distribution.

Field within the conductor – zero(free charges screen the external field)

Any excess charge resides on thesurface

0S

E d A

Field of a charged conducting sphere

Field of a thin, uniformly charged conducting wire

Field outside the wire can only point radially outward, and, therefore, mayonly depend on the distance from the wire

0

QE d A

02E

r

- linear density of charge

Field of the uniformly charged sphere

rE03

Uniform charge within a sphere of radius r

3' r

q Qa

Q - total charge

Q

V - volume density of charge

Field of the infinitely large conducting plate

- uniform surface charge densityQ

A

02E

Charges on Conductors

Field within conductor E=0

Experimental Testing of the Gauss’s Law