Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Physics for IIT JAM, UGC - NET, CSIR NET

Created by: Akhilesh Thakur

Physics : Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

The document Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
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Example: The Electrostatic Fields of a Coaxial Line 

A common form of a transmission line is the coaxial cable. 

 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

 The coax has an outer diameter b, and an inner diameter a.
The space between the conductors is filled with dielectric material of permittivity ε .

Say a voltage V0 is placed across the conductors, such that the electric potential of the outer conductor is zero, and the electric potential of the inner conductor is V0.  

The potential difference between the inner and outer conductor is therefore V0 – 0 = Vvolts. 

Q: What electric potential field VBoundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev, electric field EBoundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev and charge density ρsBoundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev is produced by this situation?

A: We must solve a boundary-value problem!  We must find solutions that:

a) Satisfy the differential equations of electrostatics (e.g., Poisson’s, Gauss’s).  

b)  Satisfy the electrostatic boundary conditions. 

Yikes! Where do we start ?

We might start with the electric potential field VBoundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev , since it is a scalar field.

a)  The electric potential function must satisfy Poisson’s equation: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

b)  It must also satisfy the boundary conditions: 

V (ρ = a) =V            V (ρ = b) = 0 

Consider first the dielectric region (a < ρ < b ).  Since the region is a dielectric, there is no free charge, and: 

ρvBoundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev =0

Therefore, Poisson’s equation reduces to Laplace’s equation: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

This particular problem (i.e., coaxial line) is directly solvable because the structure is cylindrically symmetric.  Rotating the coax around the z-axis (i.e., in the Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRevdirection) does not change the geometry at all. As a result, we know that the electric potential field is a function of ρ only ! I.E.,: 

V Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev = V ( ρ )
This make the problem much easier.  Laplace’s equation becomes: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Integrating both sides of the resulting equation, we find:

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

where C1 is some constant.  

Rearranging the above equation, we find: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Integrating both sides again, we get: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

We find that this final equation (V ( ρ ) = C 1 ln [ρ ] +C) will satisfy Laplace’s equation  (try it!).  

We must now apply the boundary conditions to determine the value of constants C1 and C2.  

  • We know that on the outer surface of the inner conductor (i.e., ρ = a ), the electric potential is equal to V0 (i.e., V ( ρ = a)=V 0 ).  
  •  And, we know that on the inner surface of the outer conductor (i.e., ρ = b ) the electric potential is equal to zero (i.e., V ( ρ = b)= 0 ).

Therefore, we can write: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Two equations and two unknowns (Cand C2)!

Solving for C1 and C2 we get:  

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

and therefore, the electric potential field within the dielectric is found to be: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Before we move on, we should do a sanity check to make sure we have done everything correctly.  Evaluating our result at ρ = a , we get: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Likewise, we evaluate our result at ρ = b : 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Our result is correct!

Now, we can determine the electric field within the dielectric by taking the gradient of the electric potential field: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Note that electric flux density is therefore: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Finally, we need to determine the charge density that actually created these fields! 

Q1: Just where is this charge? After all, the dielectric (if it is perfect) will contain no free charge.

A1:  The free charge, as we might expect, is in the conductors.  Specifically, the charge is located at the surface of the conductor.

Q2: Just how do we determine this surface charge ρs Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev ?

A2: Apply the boundary conditions! 

Recall that we found that at a conductor/dielectric interface, the surface charge density on the conductor is related to the electric flux density in the dielectric as:

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

First, we find that the electric flux density on the surface of the inner conductor (i.e., at ρ = a ) is: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

For every point on outer surface of the inner conductor, we find that the unit vector normal to the conductor is: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Therefore, we find that the surface charge density on the outer surface of the inner conductor is:

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Likewise, we find the unit vector normal to the inner surface of the outer conductor is (do you see why?): 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Therefore, evaluating the electric flux density on the inner surface of the outer conductor (i.e., ρ = b ), we find: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Note the charge on the outer conductor is negative, while that of the inner conductor is positive.  Hence, the electric field points from the inner conductor to the outer. 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

We should note several things about these solutions: 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

3) DBoundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev and EBoundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev are normal to the surface of the conductor (i.e., their tangential components are equal to zero). 

4)  The electric field is precisely the same as that given by eq. 4.31 in section 4-5! 

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Boundary Value Problems(Part- 2) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

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