Lecture 4
Electric Potential
In this lecture you will learn:
• Electric Scalar Potential
• Laplace’s and Poisson’s Equation
• Potential of Some Simple Charge Distributions
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
Conservative or Irrotational Fields
Irrotational or Conservative Fields:
r
r
Vector fields F for which ∇ × F = 0 are called “irrotational” or “conservative” fields
r
• This implies that the line integral of F
around any closed loop is zero
r r
∫ F . ds = 0
Equations of Electrostatics:
Recall the equations of electrostatics from a previous lecture:
r
∇. ε o E = ρ
r
∇×E = 0
⇒ In electrostatics or electroquasistatics, the E-field is conservative or irrotational
(But this is not true in electrodynamics)
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
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Conservative or Irrotational Fields
More on Irrotational or Conservative Fields:
r
• If the line integral of F
around any closed loop is zero …..
r r
∫ F . ds = 0
r
…. then the line integral of F between any two points is independent of any specific
Path (i.e. the line integral is the same for all possible paths between the two points)
r r
∫ F . ds = 0
r
r2
path A
r
r1
path B
r
r
r
r
r
r
⇒
⎛ r2 r r ⎞
⎛ r1 r r ⎞
⎜ ∫ F . ds ⎟
+ ⎜ r∫ F . ds ⎟
=0
r
⎜r
⎟
⎜
⎟
⎝ 1
⎠path A ⎝ r2
⎠path B
⇒
⎛ r2 r r ⎞
⎛ r2 r r ⎞
⎜ ∫ F . ds ⎟
− ⎜ r∫ F . ds ⎟
=0
⎜ rr
⎟
⎜
⎟
⎝ 1
⎠path A ⎝ r1
⎠path B
⇒
⎛ r2 r r ⎞
⎛ r2 r r ⎞
⎜ ∫ F . ds ⎟
⎜ ∫ F . ds ⎟
=
⎜ rr
⎟
⎜r
⎟
⎝ 1
⎠path A ⎝ r1
⎠path B
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
The Electric Scalar Potential - I
The scalar potential:
Any conservative field can always be written (up to a constant) as the gradient
of some scalar quantity. This holds because the curl of a gradient is always
r
zero.
If F = ∇ϕ
( )
r
Then ∇ × F = ∇ × (∇ϕ ) = 0
For the conservative E-field one writes:
(The –ve sign is just a convention)
r
E = −∇ φ
Where φ is the scalar electric potential
The scalar potential is defined only up to a constant
r
φ (r ) gives a certain electric field then the scalar
If the scalar potential
r
potential φ (r ) + c will also give the same electric field (where c is a constant)
The absolute value of potential in a problem is generally fixed by some
physical reasoning that essentially fixes the value of the constant c
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
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The Electric Scalar Potential - II
We know that:
r
E = −∇ φ
This immediately suggests that:
• The line integral of E-field between any two points is the difference of
the potentials at those points
r
r2
r∫
r1
r
r2
r r
r
r
r
E . ds = r∫ (− ∇φ ) . ds = φ (r1 ) − φ (r2 )
r
r2
r
r1
r1
• The line integral of E-field around a closed loop is zero
r r
r
∫ E . ds = ∫ (− ∇φ ) . ds = 0
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
The Electric Scalar Potential of a Point Charge
Assumption: The scalar potential is assumed to have a value equal to zero at
infinity far away from any charges
Point Charge Potential
r
ds
r
r
r
E=
q
∞
q
rˆ
4πε o r 2
r
Do a line integral from infinity to the point r where the potential needs to be
determined
0
r r ∞
r
r
r
∫r E . ds = ∫r (− ∇φ ) . ds = φ (r ) − φ (∞ ) = φ (r )
∞
r
⇒
r
∞
r
r
r
∞
φ (r ) = ∫r E . ds = ∫
r
=
r
r
∞
r
r
φ (r ) = ∫r E . ds
r
q
dr
4π ε o r 2
q
4π ε o r
r
φ (r ) =
q
4π ε o r
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
3
Electric Scalar Potential and Electric Potential Energy
The electric scalar potential is the potential energy of a unit positive charge
in an electric field
r
• Electric force on a charge of q Coulombs = q E
Potential energy of a charge q at
any point in an electric field
(Lorentz Law)
Work done by the field in moving the
charge q from that point to infinity
=
∞r
r ∞ r r
r
r
Work done = ∫r F . ds = ∫r qE . ds = q [φ (r ) − φ (∞ )] = qφ (r )
r
r
Work done on unit charge =
r
r
qφ(r )
= φ(r )
q
r
⇒ P.E. of unit charge = φ (r )
r
ds
∞
r
r
⇒
r
Potential energy of a charge of q Coulombs in electric field = q φ (r )
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
Poisson’s and Laplace’s Equation
• It is not always easy to directly use Gauss’ Law and solve for the electric fields
• Need an equation for the electric potential
r
Start from: ∇. ε o E = ρ
Use:
r
E = −∇ φ
To get: ∇. ε o (− ∇φ ) = ρ
⇒
∇2 φ = −
ρ
εo
Poisson’s Equation
If the volume charge density is zero then Poisson’s equation becomes:
∇2 φ = 0
Laplace’s Equation
Poisson’s or Laplace’s equation can be solved to give the electric scalar
potential for charge distributions
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
4
Potential of a Uniformly Charged Spherical Shell - I
σ Coulombs/m2
Use the spherical coordinate system
a
For a ≤ r ≤ ∞ :
Assume a solution:
∇2 φ = 0
⇒
1 ∂ ⎛ 2 ∂φ ⎞
⎜r
⎟=0
r 2 ∂r ⎝ ∂r ⎠
r
φ (r ) =
A
+F
r
F must be 0 so that the potential is 0
at r = ∞
For 0 ≤ r ≤ a :
∇2 φ = 0
⇒
Assume solution:
1 ∂ ⎛ 2 ∂φ ⎞
⎜r
⎟=0
r 2 ∂r ⎝ ∂r ⎠
r
φ (r ) =
B
+D
r
Potential must not become infinite
at r = 0 so B must be 0
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
Potential of a Uniformly Charged Spherical Shell - II
For 0 ≤ r ≤ a
For a ≤ r ≤ ∞
r
φ (r ) = D
E r (r ) = −
A
r
A
∂φ
E r (r ) = −
=
∂r r 2
r
φ (r ) =
∂φ
=0
∂r
a
σ
Boundary conditions
We need two additional boundary conditions to determine the two unknown
coefficients A and D
(1) At r = a the potential is continuous (i.e. it is the same just inside and just
outside the charged sphere)
D=
A
a
(2) At r = a the electric field is NOT continuous. The jump in the component of
the field normal to the shell (i.e. the radial component) is related to the
surface charge density
ε o (E r out − E r
⇒
in
)=σ
⎛A
⎞
− 0⎟ = σ
⎝ a2
⎠
εo ⎜
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
5
Surface Charge Density Boundary Condition
Suppose we know the surface normal electric
field on just one side of a charge plane with a
surface charge density σ
σ
E2 = ??
E1
Question: What is the surface normal field on
the other side of the charge plane?
Solution:
• Draw a Gaussian surface in the form of a cylinder of area A
piercing the charge plane
• Total flux coming out of the surface = ε o (E2 − E1 ) A
• Total charge enclosed by the surface = σ A
ε o (E2 − E1 ) = σ
E2 = ??
E1
ε o (E2 − E1 )A = σA
⇒ ε o (E2 − E1 ) = σ
• By Gauss’ Law:
σ
A
This an extremely important result that relates surface normal
electric fields on the two sides of a charge plane with surface
charge density σ
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
Potential of a Uniformly Charged Spherical Shell - III
For 0 ≤ r ≤ a
r
φ (r ) =
(4πσ a )
2
4π ε o a
For a ≤ r ≤ ∞
(
r
4πσ a 2
φ (r ) =
4π ε o r
)
a
σ
Sketch of the Potential:
φ (r )
r
a
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
6
Potential of a Uniformly Charged Sphere a la Poisson and Laplace
In spherical co-ordinates potential can only be a function of r (not of θ or φ )
ρ Coulombs/m3
For a ≤ r ≤ ∞ :
Assume a solution:
∇2 φ = 0
⇒
1 ∂ ⎛ 2 ∂φ ⎞
⎜r
⎟=0
r 2 ∂r ⎝ ∂r ⎠
r
φ (r ) =
Work in spherical
co-ordinates
F must be 0 so that the
potential is 0 at r = ∞
For 0 ≤ r ≤ a :
∇2 φ = −
a
A
+F
r
ρ
εo
Assume solution:
ρ
1 ∂ ⎛ 2 ∂φ ⎞
⇒ 2
⎜r
⎟=−
εo
r ∂r ⎝ ∂r ⎠
r
φ (r ) =
B
+ D +Cr2
r
particular
solution
homogenous
parts
C=−
By substituting the solution in the Poisson equation find C
ρ
6 εo
• Potential must not become infinite at r = 0 so B must be 0
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
Potential of a Uniformly Charged Sphere a la Poisson and Laplace
ρ
For 0 ≤ r ≤ a
For a ≤ r ≤ ∞
r
φ (r ) = D −
ρ
6 εo
r2
r
φ (r ) =
A
r
a
Boundary conditions
We need two additional boundary conditions to determine the two unknown
coefficients A and D
(1) At r = a the potential is continuous (i.e. it is the same just
inside and just outside the charged sphere)
(2) At r = a the radial electric field is continuous (i.e. it is the
same just inside and just outside the charged sphere)
(1) gives:
D−
(2) gives:
ρ
6 εo
a2 =
A
ρ
a= 2
3 εo
a
A
a
A=
⇒
D=
ρ
3 εo
ρ
2 εo
Er = −
∂φ
∂r
a3
a2
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
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Potential of a Uniformly Charged Sphere a la Poisson and Laplace
ρ
For 0 ≤ r ≤ a
For a ≤ r ≤ ∞
a
⎛ 4
3⎞
⎜ρ π a ⎟
r
3
⎝
⎠
φ (r ) =
4π ε o r
r
ρ ⎛ 2 r2 ⎞
⎜ a − ⎟⎟
φ (r ) =
2 ε o ⎜⎝
3⎠
Sketch of the Potential:
φ (r )
r
a
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
The Principle of Superposition for the Electric Potential
Poisson equation is LINEAR and allows for the superposition principle to hold
• Suppose for some charge density ρ1 one has found the potential φ1
• Suppose for some other charge density ρ 2 one has found the potential φ2
The superposition principle says that the sum (φ1 + φ2 ) is the solution for the
charge density ( ρ1 + ρ 2 )
A Simple Proof
∇ 2 φ1 = −
ρ1
εo
+
∇ 2 φ2 = −
ρ2
εo
=
∇ 2 (φ1 + φ2 ) = −
( ρ1 + ρ 2 )
εo
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
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Potential of a Charge Dipole
Consider Two Equal and Opposite Charges
z
P
We are interested in the potential
at a distance r from the center of
the pair in the plane of the
charges, where r >> d
θ
+q
Work in spherical co-ordinates
r+
r
r−
d
Potential contributions from the
two charges can be added
algebraically
r
φ (r ) =
d
cos(θ )
2
d
r− = r + cos(θ )
2
r+ = r −
d
cos(θ )
2
−q
q
q
−
4π ε o r+ 4π ε o r−
q
q
−
d
d
⎛
⎞
⎛
⎞
4π ε o ⎜ r − cos(θ ) ⎟ 4π ε o ⎜ r + cos(θ ) ⎟
2
2
⎝
⎠
⎝
⎠
qd
≈
cos(θ )
4π ε o r 2
=
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
Field of a Charge Dipole
r
φ (r ) ≈
qd
4π ε o r 2
cos(θ )
r
r
E = −∇ φ (r )
qd
≈
2 cos(θ ) rˆ + sin(θ )θˆ
4πε o r 3
(
)
+q
−q
Same result for the E-field was obtained
in the previous lecture by superposing
the individual E-fields (rather than the
potentials) of the two charges
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
9
Potential of a Line Charge
y
Consider an infinite line charge coming
out of the plane of slide
• The electric field, by symmetry, has only a
radial component
λ
Coulombs/m
• Draw a Gaussian surface in the form of a cylinder of
radius r and Length L perpendicular to the slide
Work in cylindrical
co-ordinates
ε oEr (2π r L ) = λ L
λ
⇒ Er =
2π ε o r
Using Gauss’ Law:
But E r = −∇ φ = −
∂φ
∂r
⇒
x
r
r
∂ φ (r )
λ
=−
∂r
2π ε o r
Upon integrating from ro to r we get: φ (r ) − φ (ro ) =
λ
2π ε o
⎛r ⎞
ln⎜ o ⎟
⎝r ⎠
Where ro is a constant of integration and is some point where the potential is
known
The problem is that this solution becomes infinite at r = ∞
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
Potential of a Line rDipole
Consider two infinite equal and opposite
line charges coming out of the plane of
slide
+λ
r
y
r+
r−
Coulombs/m
d
−λ
Coulombs/m
x
Using superposition, the potential can be written as:
r
⎛r ⎞
⎛r ⎞
λ
ln⎜⎜ o ⎟⎟ −
ln⎜⎜ o ⎟⎟
r
π
ε
2
⎝ +⎠
⎝ r− ⎠
o
⎛r ⎞
λ
=
ln⎜⎜ − ⎟⎟
2π ε o ⎝ r+ ⎠
φ (r ) =
λ
2π ε o
The final answer does not
depend on the parameter ro
Question: where is the zero of potential?
Points for which r+ equals r- have zero potential. These points constitute
the entire y-z plane
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
10
The 3D Superposition Integral for the Potential
In the most general scenario, one has to solve the Poisson equation:
r
r
ρ (r )
∇ 2 φ (r ) = −
r r
r − r'
εo
r
r'
We know that the solution for a point charge
sitting at the origin:
r
φ (r ) =
r
r
q
4π ε o r
ρ
To find the potential at any point one can sum up the contributions from
different portions of a charge distribution treating each as a point charge
r
φ (r ) = ∫∫∫
r
ρ (r ')
r r dV '
4π ε o r − r '
dV ' = dx ' dy ' dz '
r
r
Check: For a point charge at the origin ρ (r ' ) = q δ 3 (r ' ) = q δ ( x ' ) δ ( y ' ) δ (z ' )
r
φ (r ) = ∫∫∫
r
r
ρ (r ')
q δ 3 (r ')
q
q
r r dV ' =
r =
r r dV ' = ∫∫∫
4π ε o r − r '
4π ε o r − r '
4π ε o r
4π ε o r
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
ECE 303 – Fall 2006 – Farhan Rana – Cornell University
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