L05_Flux_post.pdf

7/30/2019 L05_Flux_post.pdf

1/18

LECTURE 5 slide 1

Lecture 5

Electric Flux and Flux Density,

Gauss Law

Sections: 3.1, 3.2, 3.3

Homework: D3.1, D3.2, D3.3, D3.4, D3.5; 3.3, 3.4, 3.6, 3.9,

3.13


2/18

LECTURE 5 slide 2

charge transfer from inner to outer sphere

, CQ =

Faradays Experiment (1837)

electric induction: charge deposition

without contact

displacement flux electric flux

observations

charge on outer sphere is of the same

magnitude but opposite sign

charge on outer sphere is the sameregardless of the insulating material used

flux is determined by the charge and does not depend on the medium

Q+

Q

E

insulator

b a

charge on outer electrode is the same

regardless of electrodes shape


3/18

LECTURE 5 slide 3

Flux Density and Flux 1

charge density on inner sphere > charge density on outer sphere

in ou

2 2

4 4s s

Q Q

a b

= =

flux has direction and density corresponding to the chargepolarity and density

s nS S

Q ds D ds= =

inner spherein

2 in

2 24 ( )

4 4

r r s

QD D a

a

a

a

= = = =

outer sphereou

2 ou

2 2

| |4 ( ) | |

4 4r r sb

b

QD b D

b

= = = =

n rD D=

S

ab


4/18

LECTURE 5 slide 4

Flux Density and Flux 2

at any distance r(flux density of point charge)

2

2 2

14

4 4r r r

Q QD r D

r r

= = = D a

n rD D=

S

abrcompare with E of a point charge

2

1

4r

Q

r= E a

E depends on the permittivity, D does not: D describes the

sources regardless of the medium

0=D Ein vacuum

the principle of superposition applies to D as well

2

1

4

v R

v

dvR

=

aD


5/18

LECTURE 5 slide 5

Total Flux through Closed Surface

flux through closed surface depends neither on the surface shape noron the mutual positioning of the charge and its closing surface

it depends solely on the enclosed charge

analogies: water flow, electrical current, etc.

find the mathematical expression for flux so that the above is true

( , ) ?S =D

consider a point charge at the origin and its electric flux through:

(a) sphere, (b) any surface

if the expression is true for a point charge it will be true for any

collection of charges as per the superposition principle


6/18


7/18

LECTURE 5 slide 7

Solid Angle and 1 Steradian

1 radian cuts out an arc of

length equal to the radius

1 steradian cuts out an area from a

sphere equal to (radius)2

arc

l r =

2s r=


8/18

LECTURE 5 slide 8

Differential Flux of Point Charge

differential flux through surface element of an arbitrary surface

sphere other cosds ds =

we want sphere other d d d Qd = =

sphere sphere other other cosrr nd D ds ds d D D s d = = = = D s

general expression for differential flux (independent of local surface

orientation)

d d = D s

spheredsotherds

rana

0 d


9/18

LECTURE 5 slide 9

Total Flux of Point Charge

Q

d

otherS

P

P

d

sphereS

regardless of the chosen surface

S

d Q = = D s

as per superposition principle, resultholds for any collection of charges

1Q

otherS

P

P

1S

2Q

2S

1 2Q Q Q= +

1 2 = +


10/18

LECTURE 5 slide 10

Gauss Law

flux equalsenclosed charge

no enclosed charge,

flux zero

the electric flux over a closed surface is equal to the totalcharge enclosed by the surface

v

S v

d Q dv = = = D s

Q

d

P

S

Pnana

d d Qd =

( )PD

( )PD

0d < 0d >


11/18

LECTURE 5 slide 11

Gauss Law: Applications

Gauss law makes solutions to problems with planar, cylindrical

or spherical symmetry easy

choose integration surface so that

D is everywhere either normal or tangential to surface

normal: ; tangential: 0d Dds d = =D s D s

when normal to surface, D is also constant on surface

S S

d D ds D S = = D s


12/18

LECTURE 5 slide 12

l

S

d Q l = = D s

2

0 0

2l

l

z

D d dz D l l

= =

= =

due to symmetry D =D a

, V/m2

l

= =

DE a

This result was already obtain in Lecture 4 by the superposition

principle.

Gauss Law Applications: Field of Infinite Line Charge

2

, C/m2

l

=D a

S

z

yx

d d dz =s al

l


13/18

LECTURE 5 slide 13

Gauss Law Applications: Field of Coaxial Cable

2

for

, C/m2

l

a b

=D a

problem has cylindrical symmetry

Gaussian surface chosen as cylinder of radius

solution analogous to that of line charge

for , 0b > =D

Prove that

for 0 ,2

va D

=

Homework:


14/18

LECTURE 5 slide 14

s

S

d Q A = = D s

top bottom sides

0

z z sD dxdy D dxdy d A + = D s

due to symmetry: top bottom 2z z sD D D DA A= = =

2 2

s sz z

= = =

DD a E a

Gauss Law Applications: Field of Sheet Charge

This result was already obtain in Lecture 4 by the superposition

principle.

S

x yl l=

D

xyz

na

yl

l

na

s

D


15/18

LECTURE 5 slide 15

A sphere of radius a has uniformly distributed charge of

volume density v, C/m3. Determine the electric flux density

in the sphere and out of it.

spherical symmetry of the source implies spherical symmetry offield

(1) inside the sphere

( ) ( )

( ) vS R v R

d Q R dv = = D s

2 34( ) 43

r vD R R R = ( )3

vrD R R

=

Gauss Law Applications: Field of Spherical Charge

choose integration surface as sphere


16/18

LECTURE 5 slide 16

(2) outside the sphere

3

2 3

3

2 2

4

3

4( ) 4

3

( ) 34

v v

S v

r v

v

r

d Q dv a

D R R Q a

Q a

D R R R

= = =

= =

= =

D s

rD

3

va

0

21/

a


outside the sphere field is the same as that of a point charge


17/18

LECTURE 5 slide 17


E-field magnitude in 3-D space


18/18

LECTURE 5 slide 18

You have learned about:

the flux density vectorD and how it relates to the charge Q andthe E vector

Gauss law of electrostatics

the application of Gauss law to the solution of symmetrical

problems

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