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

ji

ijjiijjiij

ij2

ij

ji

o

ji3

ji

ji

o

ij

r

r

qq

4

1qq

4

1

rr

rr

rrrrrr

rrr

rr

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

ij2

ij

j

o

ii

r

q

4

1

q rF ˆ

πε

( ) ∑≠

==

ij

ij2

ij

j

oi

i

ii

r

q

4

1

q

r

F

rE 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

Cm2

s-1

da1

da2

dS

dS = da1 x da2

|dS| = |da1| |da2|sin(π/2)

Ψ

dS`

∫ Ssurfaceclosed

.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

ϕ

ϕϕ

ˆˆˆ

ˆ

ˆ

ˆ

θn

naaS

a

θa

x

ddθsinθrdxdd

dsinθrd

dθrd

2

21

2

1

=

==

=

=

o

oo

2

2

o

q

.d

d

4

q

ddθsinθ

4

q

1ddθsinθr.

r4

q

.d

ε

πε

ϕ

πε

ϕ

πε

∫ =

Ω==

==

S

SE

n.rn

r

SE ˆˆˆ

ˆ

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

i

i

o

21

q

.d

d

4

qq

.d

ε

πε

∑

∫ =

Ω

+

=

S

SE

SE

Point charges: qi enclosed by S

q1

q2

vwithinchargetotal)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(

.d

ε

ρ∫

∫ =

rr

SE

S

∫∫

∫∫

=∇

∇=

VV

V

)dv(

1

dv.

dv.d

rE

ESE.

ρ

εo

S

oε

ρ )(

)(.

r

rE =∇

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 ε

σ

ε

σ

ε 2

=⇒=⇒=∫ ESE.

S

.dA

E.2dA

Q

d 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

A

B

q1

q

r

r1

r2

E

dl

θ

∫

∫

−=

−−=

−=

−=

B

A

21o

1

r

r 2

o

1

2

o

1

.dq

r

1

r

1

4

q

q

dr

r

1

4

q

qW

dr

r

1

4

q

qdW

2

1

lE

πε

πε

πε

0=∫ pathclosedany

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

r

1

4

q

)(

o

1

πε

φ =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

2

o

1

B

r

1

4

q

)(

dr

r

1

4

q

.d

-)()(-)(

B

πε

φ

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

3

o

1

r4

q r

E

πε

=

r

1

4

q

r

o

1

B

πε

φ =)(

3

o

1

B

r4

q

r

r

πε

φ −=∇ )(

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

1

2

r

1q

πε

ϕ

4

=

O

q1

q2

r1 r2

r12

r3

r13

r23

23o

2

13o

1

3

r

1q

r

1q

πεπε

ϕ

44

+=

∑ ∑∑ ∑ ≠<

==

ji j ij

j

i

ji j ij

j

i

r

q

q

1

2

1

r

q

q

1

W

oo πεπε 44

Electrostatic energy of charge

distribution

• For a continuous distribution

∫∫

∫

∫

−

=

−

=

=

spaceallspaceallo

spaceallo

spaceall

)(

d)(d

4

1

2

1

W

)(

d

4

1

)(

)()(d

2

1

W

r’r

r’

r’rr

r’r

r’

r’r

rrr

ρ

ρ

πε

ρ

πε

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.∇ψ

Exercise: write ψ = φ and F = ∇φ to show:

∫∫ ∇−==∴

∇−=⇒−=∇⇒

−∇==∇

v

2o

v

2

o

o

2

o

dv

2

dv

2

1W

and.

φφ

ε

ρφ

φερ

ε

ρ

φ

φ

ε

ρ

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(DivergencintegralvolumereplacesintegralSurface

identityfirstsGreen’dv.d

2

dvdv.

2

W

v

2o

v

2

v

o

∇−∇−=

∇−∇∇−=

∫∫

∫∫

φφφ

ε

φφφ

ε

S

S

( ) ∫∫ =∇=

spaceall

2o

spaceall

2o

dvE

2

dv

2

W

ε

φ

ε

)(E

2dv

dW

)( 2o

E rr

ε

ρ ==