Lectures 3,4 (Ch. 22)Gauss’s law

E

1.Flux of2. Gauss’s law3. Applicationa)Finding of Fluxb)Finding of Chargec)Conductors propertiesd)Finding of E for symmetric charge distributions(sphere, cylinder, plane)

Flux of a uniform E

EAAEEACosAE

EonAorAonEprojecteither

should cross the surface to produce a flux

n

n

n

E

Compare to a flux of water

should cross the surface to produce a fluxv

A

A

A

?EAorEA 0 EAEAtotal

For the closed surface A

is out of the surface

General definition of flux of E

E

AdEd

E AdE

enclosedQnoconstN

NNif

NEAA

NE

outin

0

~~

d

Flux of E through the closed surface containing a point charge, q

2. Arbitrary surface with arbitrary q position

1. Sphere with q placed in the center

0

22

0

44

qr

r

qEAAdE

02

0

0

22

02

0

4

444

q

r

dAqAdEor

qR

R

qdA

R

q

22 4

)(

4

)()()(

R

RqdA

r

rqdArdArEAdE

AdE

Flux of E through the closed surface containing an ensemble of charges

q1 q2

qN

Superposition principle:

N

i

iN

ii

N

ii

N

ii

QqAdEAdEAdE

1 00111

N

iiEE

1

Charges outside the closed surface do not contribute to the flux through this surface

inAd

outAd For each solid angle flux in =-flux out or

densityechvolumeais

dVQ

AdElawsGauss volumeencl

surface

arg

:'00

0 outin NN

Gauss’s Law may be used 1) to find a flux 2) Q enclosed 3) E for some symmetric charge distributions.

Figure 22.15

Find the flux through the closed surfaces: A,B, C, D.

Figure 22.16

Find the flux through the indicated closed surfaces.

Find the flux through each side.

q

Find the charge inside

Q=?

E0=500N/CEL=400N/C

=10m

Conductors properties

1.E=0 within conductor in electrostatics(otherwise F=qE and hence the charges

would be moving, i.e. nonstatic)

2.q =0 in every point inside solid conductor. It follows from the Gauss’s law, indeed, since E=0

one has

q=0 0

0enclq

AdE

Shrinking a Gaussian surface to any point

inside a conductor one gets 0enclq

Conductors properties3. E near the surface

AEALaw

sGaussfromfollowsitEb

movingbewould

electronsotherwiseEEei

surfaceconductoratoEa

0

0

:

')

)

(..

,)

Is the surface charge density, i.e. a charge per unit area

E

Conductors properties

4.In the absence of external charge placed into the cavitya) inside a cavity E=0b) there is no charge on the inner surface of the conductor(“Scooping out theorem” in electrically neutral material)

-

-

++

Faraday’s cageInsert uncharged conductor into E

E=0, qinner=0

Is it safe to be in a carduring the thunderstorm?

☺

Conductors properties

5. In the presence of external charge q inside the cavitya)Charge –q is induced on the inner surface of the cavity. It follows from the Gauss’s law. Since E=0 inside a conductor one gets:

0

0enclq

AdE

Hence qqei

qqq

inner

innerencl

..

0

b) Charge qqc resides on the outer surface of a cavity.

It follows from the charge conservation law: qqqqqqqq cinnercoutercinnerouter

Faraday’s ice-pail experiment

Calculation of E using the Gauss’s lawThree types of the symmetry for a charge distribution

1. Spherical A=4 π r2, V=4 π r3/3

1. Cylindrical Aside=2 π rL, V= π r2L

2. Plane A=L1L2, V=AL3

Application of Gauss’s law• Look for the symmetry of charge distribution• Use the Gaussian surface of the same symmetry

• Then you’ll get A

qE

qEAAdE enclencl

00

lqorAqorVq

ondestributiechuniformafor

dlqordAqordVq

enclenclencl

enclenclencl

arg

here

volume

surfacelinear

charge density

1. r<R, E=0(conductor)

2. r>R, 0

24

qrE

2r

kqE

q

Dielectrcic sphere

03

0

03

32

3

0

32

34

4

3/4

3

44:.1

r

R

QrE

R

QrrE

R

Q

rrERr

Q

2. r>R, 0

24

qrE

2r

kqE

Q

L

Q

L is the total length of the line

sheetA

Q

Q

Conducting spherical layer

a

b

Q

q

2

2

)(.3

0.2

.1

r

qQkEbr

conductorEbrar

kqEar

q-q Q+q

1. Find a surface charge density on inner and outer sides.2.Find E(r).

2

2

4

4

b

qQa

q

b

a

Dielectric spherical layer

q

Q a

b

)(

)((

3/)(4

3/)(44.3

)(.2;.1

332

33

2

33

0

332

22

abr

arkQ

r

kqE

ab

Q

arqrE

r

qQkEbr

r

kqEar

Earnshaw’s Theorem, 1842 • A collection of point charges cannot be maintained

in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges.

• Suppose a charge q is at stable equilibrium in p. P. • It means if it is moved away from this point in any

direction there should be a restoring force (opposite to the direction of a displacement.)

• It means that there is a flux of E through the nearby surface surrounding p. P.

• According to Gauss’s theorem p. P should be occupied with a charge.

• But we supposed that the charge was shifted from this point.

P

J.J. Thomson’s “plum pudding” model of atom

Sir J.J.Tomson (1856 -1940) Nobel prize for discovery of e,1906

Such atom would be unstable!

Long conducting cylinder2a

1.r<a E=0 (conductor)2.r> a the same as for theline of charge

aL

aL

L

Q 22

Connection between the linear and surface charge density

r

kE

2

Long dielectric cylinder2a

1.r>a the same result as for conducting cylinder:

20

2

2

0

2

22

2

.2

a

rk

La

QrE

La

Q

lrrlE

ar

r

kE

2

Conducting cylindrical shell with a line of charge

a b

1 2

1 21

NB: linear charge density on an inner surface of a shell ison an outer surface

1.r<a the same result as for a line in the center:

r

kE

)(2 21

2.a<r<b E=0 (conductor)

3. r>b

1

21

r

kE 12

Dielectric cylindrical shell with a line of charge

a b

1 2

1.r<a and 3.r>b the same results as for conducting cylindrical shell

2.a<r<b

)(

)(22

)()(

)(2

22

2221

222

222

0

22

0

1

abr

ark

r

kE

ababL

Q

larlrlE

A conducting slab

x

z

ya

-a

L1

L2

L2 >>a,L1>>a, r<< L2 ,r<<L1

1.|x|<a E=0 (conductor)

2.|x|>a

2EA=2σA/ε0→E= σ/ε0

σ =Q/2S, S= L1 L2

E=Q/2ε0S

A dielectric slab

x

z

ya

-a

L1

L2

L2 >>a,L1>>a, r<< L2 ,r<<L1

1.|x|>a2EA=σ*A/ε0→E= σ*/2ε0

NB: σ* =Q/S, S= L1 L2

E=Q/2ε0S the same as for conductor

2.|x|<a 2EA=ρA2x/ε0

ρ=Q/S2a= σ*/2a

E=σ*x/2aε0

0

Conclusions1. Inside of charged bodiesConductors E=0Dielectrics E ~ rIndependent on the symmetry of charge distribution

2.Outside of charged bodiesE is the same for conductors and dielectrics.It depends on the symmetry of charge distribution:Sphere E ~ 1/r2

Cylinder E ~1/r (r<<L)Plane E=const (r<< L1 and L2)

3. For finite size bodies at r→∞ E=kQtotal/r2

4. E jumps on the surface of the charged conductor

E changes continuously on the surface of dielectric