Electric field generated by an electric...

(22-7)

We will determine the electric field generated by the electric dipole shown in the figure using the principle of superposition. The positive charge generates at P an electric

field whose magnitu

Er

( ) 2

( ) 2

1de 4

The negative charge creates an electric field 1with magnitude

4

o

o

qEr

qEr

πε

πε

++

−−

=

=

( ) 21 1 2x x−+ −;

Electric field generated by an electric dipole

( ) ( )

( ) ( )

2 2

2 2

2 2

2

2

The net electric field at P is:

1 4

1 4 / 2 / 2

1 1 4 2 2

We assume: 12

1 14

o

o

o

o

E E E

q qEr r

q qz d z d

q d dEz z z

dz

q d dEz z z

πε

πε

πε

πε

+ −

+ −

− −

= −

⎛ ⎞= −⎜ ⎟

⎝ ⎠⎛ ⎞

= −⎜ ⎟⎜ ⎟− +⎝ ⎠⎡ ⎤⎛ ⎞ ⎛ ⎞= − − +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

=

3 3

12 2o o

qd pz zπε πε

= =

(22-7)

( ) 21 1 2x x−+ −;

Electric dipole: Two charges that are equal in magnitude but of opposite signal .

( ) ?=rEρρ+rρ

rρ

−rρ

q+

q−

o

2/d

2/d

dr>>

( ) ( ) ( )

( )

2 2 3 3

2 23

3 233 2

332 2 22

2 22 2

32

ˆ ˆ

22 4

14 4

31 , 2

d d

k q k qkq kqE r r r r rr r r r

r r r r

d d dr r r r r r r

d r d dr r d rr r

r d rrr

+ + −+ − + −

+ −

−−−−

+ + + +

−−

−

− −= + = +

= − = +

⎛ ⎞⋅= = ⋅ = ⋅ − ⋅ +⎜ ⎟

⎝ ⎠

⎡ ⎤⎛ ⎞⎛ ⎞ ⋅= − ⋅ + = − +⎢ ⎥⎜ ⎟⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎛ ⎞⋅ ⋅

≈ +⎜ ⎟⎝ ⎠

r r

r r r r

r r r r

r rr r r r r

rrrr

r rr r

( )

( )

2 1

2 2

333 322 3 2 3 3

3 32 2

1 14

3 1ˆ ˆ1 3 3( ) ( )2

2 31 13 2 2 2

xnd d x n xr r

r d kq r d kqr r r r r r d d r r dr r r r r

r d d r d dE kq r r kq r rr r

>>Δ

−−− −− − − −

− −

>> + Δ ≈ + Δ

⎛ ⎞ ⎡ ⎤⋅ ⋅ ⎡ ⎤= = ⋅ = − = − = ⋅ − ∝⎜ ⎟ ⎢ ⎥ ⎣ ⎦⎝ ⎠ ⎣ ⎦

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛⋅ ⋅= ⋅ + − − ⋅ − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝

r rr r r rvr r r

r r r rr rr r r

⎞⎟⎠

Dipole moment : p qd=rv

( )

[ ]

[ ]

3 3

3 3 3

3 3 3

1.

2. / / , 0 ˆˆ ˆ 3 2 2

3. / / , 0 ˆ ˆˆ ˆ ˆ 3( ) 2 2 , ( )

kq kpr d E dr r

r d r d kq kq kpE dr dr dr r r

r d r dkq kq kpE d r dr d r dr r r

⊥ ⇒ = − = −

⋅ >

⎡ ⎤⇒ = − = =⎣ ⎦

⋅ <

⎡ ⎤⇒ = − − = = = −⎣ ⎦

vr rrr

r vr v

r v

r vr v

r v

Effect of dielectric material

( ) vacuum: 00 kε

20

0 rqk

E free=

freeqinducedEρ

freeq

boundq

000 2 2 2

0

14

free free freeinduced

r r r

k q q kqEE E Er r rε ε πε ε

= − = = = =

Consider the electric dipole shown in the figure in the presence of a uniform (constant magnitude and direction)

electric field along the -axis.The electric field exerts a force on the posit

E xF qE+ =

r

ive charge and a force on the negatice charge. The net force on the dipole 0net

F qE

F qE qE

− = −

= − =

(22-14)

Forces and torques exerted on electric dipoles by a uniform electric field

The net torque generated by and about the dipole center is:

sin sin sin sin2 2

In vector form: The electric dipole in a uniform electric field does not movebut

F Fd dF F qEd pE

p E

τ τ τ θ θ θ θ

τ

+ −

+ − + −= + = − − = − = −

= ×rr r

can rotate about its center.

p Eτ = ×rr r0netF =

r

(22-14)

A

B

θU

180°

90 90

90

sin

sin cos

U d pE d

U pE d pE p E

θ θ

θ

τ θ θ θ

θ θ θ

° °

°

′ ′= − = −

′= − = − = − ⋅

∫ ∫

∫rr

prEr

Er

pr

min

At point ( 0) has a minimumvalue It is a position of equilibriums

A

table

UU pE

θ == −

max

At point ( 180 ) has a maximumvalue It is a position of equilibriumunst le

B

ab

UU pE

θ = °= +

U p E= − ⋅rr

cosU pE θ= −

(22-15)

Potential energy of an electric dipole in a uniform electric field

Chapter 24

Gauss’s Law

Electric FluxElectric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the fieldΦE = EA

Electric Flux, General AreaThe electric flux is proportional to the number of electric field lines penetrating some surfaceThe field lines may make some angle θwith the perpendicular to the surfaceThen ΦE = EA cos θ

Electric Flux, Interpreting the Equation

The flux is a maximum when the surface is perpendicular to the fieldThe flux is zero when the surface is parallel to the fieldIf the field varies over the surface, 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

cosE i i i i iE A θΔΦ = Δ = ⋅ ΔE A

0surface

limi

E i iAE A d

Δ →Φ = ⋅ Δ = ⋅∑ ∫ E A

Electric Flux, finalThe surface integral means the integral must be evaluated over the surface in questionIn general, the value of the flux will depend both on the field pattern and on the surfaceThe units of electric flux will be N.m2/C2

Electric Flux, Closed SurfaceAssume a closed surfaceThe vectors ΔAipoint in different directions

At each point, they are perpendicular to the surfaceBy convention, they point outward

Active Figure 24.4

(SLIDESHOW MODE ONLY)

Flux Through Closed Surface, cont.

At (1), the field lines are crossing the surface from the inside to the outside; θ < 90o, Φ is positiveAt (2), the field lines graze surface; θ = 90o, Φ = 0At (3), the field lines are crossing the surface from the outside to the inside;180o > θ > 90o, Φ is negative

Flux Through Closed Surface, final

The net flux through the surface is proportional to the net number of lines leaving the surface

This net number of lines is the number of lines leaving the surface minus the number entering the surface

If En is the component of Eperpendicular to the surface, then

E nd E dAΦ = ⋅ =∫ ∫E A旄

Gauss’s Law, IntroductionGauss’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

PI Electrostatics CT9A cylindrical piece of insulating material is placed in an external electric field, as shown. The net electric flux passing through the surface of the cylinder is

1、positive 2、negative 3、zero