# Gauss law 1

Maxwell’s Equations Part 1: Gauss’s Law for the Electric Field
Maxwell’s Equations Part 1: Gauss’s Law for the Electric Field

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

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
• 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 ==

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