phy103_lec_9
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Semester II, 2015-16
Department of Physics, IIT Kanpur
PHY103A: Lecture # 9(Text Book: Intro to Electrodynamics by Griffiths, 3rd Ed.)
Anand Kumar Jha
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Notes
•First Problem Solving Session Today:
5-6 pm: TB-205, TB-206, TB-207
(TAs and Instructors will be around to help)
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Summary of Lecture # 8:
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• Force per unit area on a conductor: = other =
20
• The electrostatic pressure on a conductor: = 20 =
02
• Capacitance is defined as: ≡
• The work necessary to charge a capacitor upto charge : = 12 = 12
= 1 + + 0 � ⋅ • The electrostatic interaction energy:•
Applications of Electrostatics:(i) Faraday Cage
(ii) Capacitor
• Special techniques: Laplace’s Equation in one-dimension
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Questions 1:
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Does the force on depend on , if the cavity was not spherical
- Force on ? 0 - Force on ? 0
Ans: No
Questions 2:
If = , how does induced charge distribute itself on the inner surface?Ans: Within classical electrodynamics, the induced charge will distribute as usual.
In quantum electrodynamics, is the minimum charge.Induced charge density is interpreted probabilistically.
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Special Techniques: Laplace’s Equation
() =1
40 �r ̂
V =
− 0
V() =1
40 �r
Q: How to find electric field ?
Ans: (Coulomb’s Law)
Very difficult to calculate the integral except for very simple situation
Alternative:
This integral is relatively easier but in general still difficult to handle
Alternative:
When = 0 V = 0 (Laplace’s Equation)(Poisson’s Equation)
If = 0 everywhere, = 0 everywhereIf
is localized, what is
away from the charge distribution?
Express the above equation in the different form.
First calculate the electric potential
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Laplace’s Equation
V = 0
(Laplace’s Equation) (In Cartesian coordinates)
+
+
= 0
If the potential V() is a solution to the Laplace’s equation then V() is theaverage value of potential over a spherical surface of radius centered at .
As a result, V() cannot have local maxima or minima; the extreme valuesof V() must occur at the boundaries.
= 1
4
Why? Because if the potential has a maximum value V
max(
) at
then one could draw a small sphere around such that everyvalue of potential on that sphere and thus the average would
be smaller than Vmax()
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Laplace’s Equation
V = 0
(Laplace’s Equation) (In Cartesian coordinates)
+
+
= 0
If the potential V() is a solution to the Laplace’s equation then V() is theaverage value of potential over a spherical surface of radius centered at .
Proof:
=1
4 =
1
4 � � 40
01
r
sin
0
= 24 40 � + − 2cos
0 sin
=24
40 �
(
1
+ − 2cos )
0
= 24 40 + − ( − ) = 40 7
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Laplace’s Equation
V = 0
(Laplace’s Equation) (In Cartesian coordinates)
+
+
= 0
If the potential V() is a solution to the Laplace’s equation then V() is theaverage value of potential over a spherical surface of radius centered at .
=1
4 = 40 We have proved the theorem for a point charge but
since potentials follow the principle of linear
superposition, the theorem is proved for any arbitrary
charge distribution.
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QED
Proof:
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Laplace’s Equation in two dimensions
If the potential V(, ) is a solution to the Laplace’s equation then V(, ) isthe average value of potential over a circle of radius centered at (, ).
, = 12
As a result, V(, ) cannot have local maxima or minima; the extremevalues of V(
,
) must occur at the boundaries.
If the potential V() is a solution to the Laplace’s equation then V() is theaverage of the potential at
+
and
−
, = 12 + + ( − )
As a result, V() cannot have local maxima or minima; the extreme valuesof V(
) must occur at the end points.
Laplace’s Equation in one dimension
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Laplace’s Equation (Solutions without solving it)
First Uniqueness Theorem: The solution to Laplace’sEquation in some volume V is uniquely determined if
V is specified on the boundary surface .
Proof: Suppose V1 and V are two distinct solutions to Laplace’s equationwithin volume V with the same value on the boundary surface .
V
1= 0
V
= 0 and
V3 ≡ V1 − V V3 = V1 − V = V1 − V = 0 V3 also satisfies Laplaces’s equation.
What is the value of V3 at the boundary surface ?
0 (Because at the boundary, V1 = V. Hence, V3 = V1 − V=0)But Laplace’s equation does not allow for any local extrema.
So, since V3 = 0 at the boundary, V3 must be 0 everywhere.
Hence V1 = V QED 10
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Ex. 3.1 (Griffiths, 3rd Ed. ): What is the potential
inside an enclosure with no charge and surrounded
completely by a conducting material?
Potential on the cavity-wall is a constant V = V0.
Note 1: V = V0 is a solution of the Laplaces’s equation inside the cavity
Note 2: V = V0 satisfies the conditions on the boundary surface.
Therefore, V = V0 must be the solution of the problem.
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Laplace’s Equation (Solutions without solving it)
Corollary to First Uniqueness Theorem: The potential in a volume
is uniquely determined if (a) the charge desity throughout theregion and (b) the value of V at all boundaries, are specified.
Proof: Suppose V1 and V are two distinct solutions to Poisson’sequation in a region with volume V and charge density . V1 and V have the same value at the boundary surface .
V1 = −
0 and
V3 ≡ V1 − V V3 = V1 − V = V1 − V = 0 V3 satisfies Laplaces’s equation.
What is the value of V3 at the boundary surface ?
0 (Because at the boundary, V1 = V. Hence, V3 = V1 − V=0)But Laplace’s equation does not allow for any local extrema.
So, if V3 = 0 at the boundary, it must be 0 everywhere.
Hence V1 = V
V = −
0
QED 12
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Laplace’s Equation (Solutions without solving it)
Second Uniqueness Theorem: In a volume V
surrounded by conductors and containing a
specified charge density
, the electric field is
uniquely determined if the total charge on eachconductor is given.
⋅ = 0
Proof: Suppose and are two distinctelectric fields satisfying the above conditions.
⋅ = 0 and
≡ −
(Gauss’s law in space between the conductors)
⋅
i
=0 ⋅
i
=0 and
(for Gaussian surface
enclosing ith conductor)
⋅
=0
⋅
= 0 and(for Gaussian surface
enclosing all conductors)
⋅ = ⋅ − = 0 ⋅
i
= 0
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Laplace’s Equation (Solutions without solving it)
Proof:
We know, V3 as well as V1 and V are allconstants over a conducting surface ⋅ (V3) = V3( ⋅ ) + ⋅ (V3) ( Using ⋅ = 0 & = −3)
(Applying divergence theorem)
= − 3 (Integrating over the entire volume)� ⋅ (V3)
V
= −� 3 V
V3 ⋅ = −� 3 V V3 ⋅
= −� 3 V
(Since V3 is a constant on the outer boundary )
0 =
−� 3
V 3 = 0 (everywhere)�
� = QED(Since
∮ ⋅
i
= 0)
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Second Uniqueness Theorem: In a volume V
surrounded by conductors and containing a
specified charge density
, the electric field is
uniquely determined if the total charge on eachconductor is given.
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Laplace’s Equation (Solutions without solving it)
Comments on Uniqueness theorem:
•If certain conditions are fulfilled, the Uniqueness theorems guaranteethat the solution is unique.
• Even if a solution is obtained by mere guess or intuition that satisfies
all the necessary conditions, it must be the unique solution.
• Uniqueness theorems do not directly help solve Laplace’s or Poisson’s
equation. They help establish that the solution is unique.
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