2). Gauss’ Law and Applications

• Coulomb’s Law: force on charge i due to charge j is

• Fij is force on i due to presence of j and acts along line of centres rij. If qi qj are same sign then repulsive force is in direction shown

• Inverse square law of force

( )

ˆ

ˆ

ji

jiijjiijjiij

ij2ij

ji

oji3

ji

ji

oij

r

r

qq

4

1qq

4

1

rr

rrrrrrrr

rrrrr

F

−−

=−=−=

=−−

=πεπε

O

ri

rj

ri-rj

qi

qj

Fij

Principle of Superposition

• Total force on one charge i is

• i.e. linear superposition of forces due to all other charges

• Test charge: one which does not influence other ‘real charges’ – samples the electric field, potential

• Electric field experienced by a test charge qi ar ri is

∑≠

=ij

ij2ij

j

oii

r

q

4

1q rF ˆ

πε

( ) ∑≠

==ij

ij2ij

j

oi

iii

r

q

4

1

qr

FrE ˆ

πε

Electric Field

• Field lines give local direction of field

• Field around positive charge directed away from charge

• Field around negative charge directed towards charge

• Principle of superposition used for field due to a dipole (+ve –ve charge combination). Which is which?

qj +ve

qj -ve

Flux of a Vector Field• Normal component of vector field transports fluid across

element of surface area

• Define surface area element as dS = da1 x da2

• Magnitude of normal component of vector field V is

V.dS = |V||dS| cos(Ψ)

• For current density j

flux through surface S is

Cm2s-1

da1

da2

dS

dS = da1 x da2

|dS| = |da1| |da2|sin(π/2)

ΨdS`

∫S surface closed

.dSj

• Electric field is vector field (c.f. fluid velocity x density)• Element of flux of electric field over closed surface E.dS

da1

da2

n

θ

φ

Flux of Electric Field

ϕ

ϕϕϕ

ˆˆˆ

ˆ

ˆ

ˆ

θn

naaS

a

θa

x

d dθ sinθ r d x dd

d sinθ rd

dθ rd

221

2

1

=

==

==

o

oo

22

o

q.d

d4

qd dθ sinθ

4

q

1 d dθ sinθr .r4

q.d

ε

πεϕ

πε

ϕπε

∫ =

Ω==

==

S

SE

n.r nr

SE ˆˆˆˆ

Gauss’ Law Integral Form

• Factors of r2 (area element) and 1/r2 (inverse square law) cancel in element of flux E.dS

• E.dS depends only on solid angle dΩ

da1

da2

n

θ

φ

Integral form of Gauss’ Law

o

ii

o

21

q.d

d4

qq.d

ε

πε

∑∫ =

Ω+=

S

SE

SE

Point charges: qi enclosed by S

q1

q2

v withincharge total)d(

)dv(

.d

V

o

V

=

=

∫

∫∫

vr

r

SE

ρ

ε

ρ

S

Charge distribution ρ(r) enclosed by S

Differential form of Gauss’ Law• Integral form

• Divergence theorem applied to field V, volume v bounded by surface S

• Divergence theorem applied to electric field E

∫∫∫ ∇==V

SS

dv .ddS. VSV. nV

V.n dS .V dv

o

V

)d(

.dε

ρ∫∫ =

rr

SES

∫∫

∫∫

=∇

∇=

VV

V

)dv(1

dv .

dv .d

rE

ESE.

ρε o

Soε

ρ )( )(.

rrE =∇

Differential form of Gauss’ Law

(Poisson’s Equation)

Apply Gauss’ Law to charge sheet

• ρ (C m-3) is the 3D charge density, many applications make use of the 2D density σ (C m-2):

• Uniform sheet of charge density σ = Q/A• By symmetry, E is perp. to sheet• Same everywhere, outwards on both sides• Surface: cylinder sides + faces • perp. to sheet, end faces of area dA • Only end faces contribute to integral

+ + + + + ++ + + + + +

+ + + + + ++ + + + + +

E

EdA

ooo εσ

εσ

ε 2=⇒=⇒=∫ ESE.

S

.dAE.2dA

Qd encl

• σ’ = Q/2A surface charge density Cm-2 (c.f. Q/A for sheet)• E 2dA = σ’ dA/εo

• E = σ’/2εo (outside left surface shown)

Apply Gauss’ Law to charged plate

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

E

dA

• E = 0 (inside metal plate)

• why??

+

+

+

+

+

+

+

+• Outside E = σ’/2εo + σ’/2εo = σ’/εo = σ/2εo

• Inside fields from opposite faces cancel

Work of moving charge in E field• FCoulomb=qE• Work done on test charge dW • dW = Fapplied.dl = -FCoulomb.dl = -qE.dl = -qEdl cos θ• dl cos θ = dr

• W is independent of the path (E is conservative field)

A

B

q1

q

rr1

r2E

dl

θ

∫

∫

−=

−−=

−=

−=

B

A

21o

1

r

r 2o

1

2o

1

.dq

r1

r1

4

qq

drr1

4

qqW

drr1

4

qqdW

2

1

lE

πε

πε

πε

0=∫path closed any

lE.d

Potential energy function

• Path independence of W leads to potential and potential energy functions

• Introduce electrostatic potential

• Work done on going from A to B = electrostatic potential energy difference

• Zero of potential energy is arbitrary– choose φ(r→∞) as zero of energy

r1

4

q)(

o

1

πεφ =r

( )

∫−=

==B

A

BA

.dq

)(-)(q)PE(-)PE(W

lE

ABAB φφ

Electrostatic potential

• Work done on test charge moving from A to B when charge q1 is at the origin

• Change in potential due to charge q1 a distance of rB from B

( )

Bo

1

r

2o

1

B

r

1

4

q)(

drr

1

4

q

.d

-)()(-)(

B

πεφ

πε

φφφ

=

−=

−=

=∞→

∫

∫

∞

∞

B

lE

BAB

0

( ))(-)(q)PE(-)PE(WBA ABAB φφ==

Electric field from electrostatic potential

• Electric field created by q1 at r = rB

• Electric potential created by q1 at rB

• Gradient of electric potential

• Electric field is therefore E= – φ

3o

1

r4

q rE

πε=

r

1

4

qr

o

1B πε

φ =)(

3o

1B r4

qr

r

πεφ −=∇ )(

Electrostatic energy of charges

In vacuum

• Potential energy of a pair of point charges

• Potential energy of a group of point charges• Potential energy of a charge distribution

In a dielectric (later)

• Potential energy of free charges

Electrostatic energy of point charges• Work to bring charge q2 to r2 from ∞ when q1 is at r1 W2 = q2 φ2

• NB q2 φ2 = q1 φ1 (Could equally well bring charge q1 from ∞)• Work to bring charge q3 to r3 from ∞ when q1 is at r1 and q2 is at

r2 W3 = q3 φ3

• Total potential energy of 3 charges = W2 + W3

• In general

O

q1 q2

r1 r2

r12

12o

12 r

1q

πεϕ

4=

O

q1 q2

r1 r2

r12

r3

r13

r23

23o

2

13o

13 r

1q

r

1q

πεπεϕ

44+=

∑ ∑∑ ∑≠<

==ji j ij

ji

ji j ij

ji r

qq

1

2

1

r

qq

1W

oo πεπε 44

Electrostatic energy of charge distribution

• For a continuous distribution

∫∫

∫

∫

−=

−=

=

space allspace allo

space allo

space all

)(d)(d

4

1

2

1W

)(d

4

1)(

)()(d2

1W

r'r

r'r'r r

r'r

r' r'r

rr r

ρρπε

ρπε

φ

φρ

Energy in vacuum in terms of E• Gauss’ law relates ρ to electric field and potential• Replace ρ in energy expression using Gauss’ law

• Expand integrand using identity:∇.ψF = ψ∇.F + F.∇ψExercise: write ψ = φ and F = ∇φ to show:

∫∫ ∇−==∴

∇−=⇒−=∇⇒

−∇==∇

v

2o

v

2o

o

2

o

dv 2

dv 21W

and.

φφερφ

φερερφ

φερ

EE

( )( ) 22

22

.

.

φφφφφ

φφφφφ

∇−∇∇=∇⇒

∇+∇=∇∇

Energy in vacuum in terms of E

For pair of point charges, contribution of surface term 1/r -1/r2 dA r2 overall -1/r

Let r → ∞ and only the volume term is non-zero

Energy density

( )

( ) ( ) ( )

theorem) e(Divergenc integral volume replaces integral Surface

identity first sGreen' dv.d2

dvdv .2

W

v

2o

v

2

v

o

∇−∇−=

∇−∇∇−=

∫∫

∫∫

φφφε

φφφε

S

S

( ) ∫∫ =∇=space all

2o

space all

2o dvE2

dv2

W εφε

)(E2dv

dW)( 2o

E rrερ ==