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Coulomb’s Law
• Coulomb’s Law: force on charge i due to
charge j is
qiq j
1 qiq j ˆ
1
(r − rj ) =
Fij =
r
3 i
2 ij
4πε o rij
4πε o r − r
i
j
rij = ri − rj
rij = ri − rj
rˆij =
ri − rj
ri − rj
• 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
ri
direction shown
• Inverse square law of force
O
Fij
qi
ri-rj
rj
qj
Principle of Superposition
• Total force on one charge i is
Fi = qi ∑
j≠i
1
qj
4πε o rij
2
rˆij
• 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
Fi
1 qj ˆ
Ei (ri ) = = ∑
r
2 ij
qi j≠i 4πε o rij
Electric Field
• Field lines give local direction of field
qj +ve
• 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
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)
da2
• For current density j
flux through surface S is
∫ j.dS Cs-1
closed surface S
dS
da1
Ψ
dS`
dS = da1 x da2
|dS| = |da1| |da2|sin(π/2)
Flux of Electric Field
• Electric field is vector field (c.f. fluid velocity x density)
• Element of flux of electric field over closed surface E.dS
da1 = r dθ θˆ
n
da 2 = r sinθ dϕ ϕˆ
da2
θ
φ
da1
dS = da1 x da 2 = r 2 sinθ dθ dϕ nˆ
nˆ = θˆ x ϕˆ
q rˆ 2
. r sinθ dθ dϕ nˆ rˆ.nˆ = 1
E.dS =
2
4πε o r
=
q
4πε o
q
∫ E.dS = ε
S
o
sinθ dθ dϕ =
q
4πε o
dΩ
Gauss’ Law Integral Form
Integral form of Gauss’ Law
• Factors of r2 (area element) and 1/r2 (inverse square law)
cancel in element of flux E.dS
q1 + q2
E.dS =
dΩ
• E.dS depends only on solid angle dΩ
4πε o
∫ E.dS =
n
da2
da1
θ
q1
φ
q2
S
∑q
i
i
εo
Point charges: qi enclosed by S
∫ ρ (r )dv
V
E
.d
S
=
∫
S
εo
∫ ρ (r )dv = total charge within v
V
Charge distribution ρ(r) enclosed by S
Differential form of Gauss’ Law
• Integral form
∫ E.dS =
∫ ρ (r )dr
V
S
εo
• Divergence theorem applied to field V, volume v bounded by
surface S ∫ V.n dS = ∫ V.dS = ∫ ∇.V dv
S
S
V
V.n dS
“.V dv
• Divergence theorem applied to electric field E
∫ E.dS = ∫ ∇.E dv
V
S
1
∫ ∇.E dv = ε ∫
V
o
V
ρ (r )dv
∇.E(r ) =
ρ (r )
εo
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
dA
E
By symmetry, E is perpendicular 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.dS =
S
Q encl
εo
⇒ E.2dA =
σ .dA
σ
⇒E =
εo
2ε o
E
Apply Gauss’ Law to charged plate
σ’ = Q/2A surface charge density Cm-2 (c.f. Q/A for sheet)
E 2dA = 2σ’ dA/εo (surface encloses whole plate)
E = σ’/εo (outside left surface shown)
E = 0 (inside metal plate)
why??
E
• Outside E = σ’/2εo + σ’/2εo = σ’/εo = σ/2εo
• Inside fields from opposite faces cancel
+
+
+
+
+
+
+
+
dA
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
•
•
•
•
•
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
dW = −q
W = −q
q1
1 dr
4πε o r 2
q1
4πε o
= −q
∫
r1
1 dr
r2
q1 ⎛ 1 1 ⎞
⎜ − ⎟
4πε o ⎝ r1 r2 ⎠
B
= −q∫ E.dl
A
•
r2
B
E
r2
θ
q
A
r
r1
dℓ
∫ E.dl = 0
any closed path
q1
W is independent of the path (electrostatic E field is conservative)
Potential energy function
• Path independence of W leads to potential and potential
energy functions
• Introduce electrostatic potential
φ (r ) =
q1 1
4πε o r
• Work done on going from A to B = electrostatic potential
energy difference
WBA = PE(B) - PE( A ) = q(φ (B) - φ ( A ))
B
• Zero of potential energy is arbitrary
– choose φ(r→∞) as zero of energy
= −q∫ E.dl
A
Electric field from electrostatic potential
q1
• 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
r
E=
4πε o r 3
q1 1
φ (rB ) =
4πε o r
q1
r
∇φ (rB ) = −
4πε o r 3
Electrostatic potential of point dipole
•
•
+/- charges, equal magnitude, q, separation a
axially symmetric potential (z axis)
2
z
q+
⎛a⎞
2
2
r± = r + ⎜ ⎟ m a r cosθ
⎝2⎠
r+
r
a/2 θ
p
a/2
q-
ϕ (r ) =
q ⎛ 1 1⎞
⎜ − ⎟
4πε o ⎝ r+ r- ⎠
2
⎛
⎞
a
a
⎞
⎛
2⎜
= r 1+ ⎜ ⎟ m cosθ ⎟
⎜ ⎝ 2r ⎠
⎟
r
⎝
⎠
rx
⎞
1 1 ⎛⎜ ⎛ a ⎞ a
= 1+ ⎜ ⎟ m cosθ ⎟
⎟
r±
r ⎜⎝ ⎝ 2r ⎠
r
⎠
1 a
= ± 2 cosθ
r 2r
qa cosθ p cosθ
ϕ (r ) =
=
2
4πε o r
4πε o r 2
2
−1
2
Electrostatic potential of point dipole
•
•
Equipotential lines for an electric dipole || z axis
Contours on which electric potential is constant
1
0.5
0
-0.5
-1
-2
-1
ϕ (r ) =
p cosθ
4πε o r 2
0
1
2
•
Equipotential lines perp. to field lines
Electrostatic potential of point dipole
•
Field lines for a point dipole || z axis
•
Generated from E = -∇φ
r
k
φ
θ
θ
j
i
⎛ ∂ ∂ ∂ ⎞
(i, j, k )
∇ Cart. = ⎜⎜ , , ⎟⎟
⎝ ∂x ∂y ∂z ⎠
⎛∂ 1 ∂
∂ ⎞ ˆ ˆ ˆ
1
⎟⎟ (r,θ ,φ )
∇ Sph.Pol. = ⎜⎜ ,
,
⎝ ∂r r ∂θ r sinθ ∂φ ⎠
φ
ϕ (r ) =
p
4πε o
cos θ
r2
E(r,θ ) = −∇ϕ (r,θ ) =
p ⎛ 2cos θ sin θ ⎞
, 3 , 0⎟
⎜
3
4πε o ⎝ r
r
⎠
← NB diagram not a point dipole
Electrostatic potential of point dipole
•
More generally, dipole direction given by p vector
ϕ (r ) =
1
4πε or
3
p.r
p = (p x , p y , p z ) r = (x, y, z)
∂ ⎛ p.r ⎞ ⎛ 1 3x 2 ⎞
3xz
3xy
⎜
⎟
p
p
pz
=
−
−
−
⎜ 3 ⎟ ⎜ 3
y
5 ⎟ x
5
5
∂x ⎝ r ⎠ ⎝ r
r
r
r ⎠
1
E(r,θ ) = −∇ϕ (x, y, z) =
T.p
4πε o
(T )ij =
3x i x j - r 2δ ij
r5
Dipole field propagation tensor
Electrostatic energy of point charges
•
Work to bring charge q2 to r2 from ∞ when q1 is at r1 W2 = q2 φ2
q1
q1
q2
r12
q2
r12
r13
r1
r2
r1
q1
1
ϕ2 =
4πε o r12
•
r2
q1
1
q2 1
ϕ3 =
+
4πε o r13 4πε o r23
r3
O
•
•
r23
O
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
1
qj
1 1
• In general W =
q
=
∑ i ∑j r 2 4πε
4πε o i< j
ij
o
qj
∑q ∑ r
i≠ j
i
j
ij
Electrostatic energy of charge
distribution
•
For a continuous distribution of charge
1
W=
dr ρ (r )φ (r )
∫
2 all space
φ (r ) =
W=
1
4πε o
1 1
2 4πε o
∫
dr'
all space
ρ (r' )
r − r'
∫ dr ρ (r )
all space
∫
all space
dr'
ρ (r' )
r − r'
Energy in vacuum in terms of E
•
•
Gauss’ law relates ρ to electric field and potential
Replace ρ in energy expression using Gauss’ law
ρ
and E = −∇φ
εo
ρ
⇒ ∇ 2φ = − ⇒ ρ = −ε o∇ 2φ
εo
ε
∴ W = 1 ∫ φ ρ dv = − o ∫ φ ∇ 2φ dv
2
∇.E =
v
•
2
v
Expand integrand using identity:
∇.ψF = ψ∇.F + F.∇ψ
2
∇
.
φ
∇
φ
=
φ
∇
φ + (∇φ )
Exercise: write ψ = φ and F = ∇φ to show:
2
⇒ φ∇ 2φ = ∇.φ∇φ − (∇φ )
2
Energy in vacuum in terms of E
εo ⎡
⎤
W = − ⎢ ∫ ∇.φ∇φ dv − ∫ (∇φ ) dv ⎥
2 ⎣v
v
⎦
⎤
εo ⎡
2
= − ⎢ ∫ (φ∇φ ).dS − ∫ (∇φ ) dv ⎥ (Green' s first identity )
2 ⎣S
v
⎦
Surface integral replaces volume integral (Divergenc e theorem)
2
For pair of point charges, contribution of surface term
f ~ 1/r “f ~ -1/r2 dA ~ r2 overall ~ -1/r
Let r → ∞ and only the volume term is non-zero
W=
Energy density
εo
2
∫ (∇φ ) dv =
2
εo
all space
dW ε o 2
= E (r )
dv
2
2
2
E
∫ dv
all space
Energy of charge distribution in
external potential
•
Energy of localised charge distribution ρ(r) in external potential φ(r)
U=
∫ dr ρ (r )φ (r )
all space
f(a + h) = f(a) + hf' (a) + ... 1- D Taylor expansion
φ (r ) = φ (0) + r.∇φ (0) + ...
U=
3 - D Taylor expansion about origin
∫ dr ρ (r )(φ (0) + r.∇φ (0) + ...)
all space
= φ (0 )
∫ dr ρ (r )
+ ∇φ (0).
all space
∫ dr ρ (r )r
+ ...
all space
U = Qφ (0) + p.∇φ (0) + ... = Qφ (0) − p.E(0) + ...
Q=
∫ dr ρ (r )
all space
p=
∫ dr ρ (r )r
all space
Energy of electric dipole in electric field Up = -p.E