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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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