Gauss' law relates the electric flux through a closed surface to the enclosed charge. It can be written in both integral and differential forms. The integral form states that the total flux is equal to the enclosed charge divided by the permittivity of free space. The differential form is Poisson's equation, which relates the divergence of the electric field to the charge density. Gauss' law can be applied to problems involving point charges, charge sheets, and continuous charge distributions. The electrostatic potential and electric field can be derived from each other using calculus operations. The potential energy of a system of charges can be expressed in terms of either the potentials or the electric fields.

1. 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

2. 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

3. 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

4. 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

5. • 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

6. • 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

7. 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)

8. 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

9. • 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

10. 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

11. 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

12. 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==

13. 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 −=∇ )(

14. 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

15. 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

16. 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

17. 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
∇−∇∇=∇⇒
∇+∇=∇∇

18. 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 ==