Boundary Value Problems(Part- 1) - 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- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

The document Boundary Value Problems(Part- 1) - 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: Dielectric Filled Parallel Plates Consider two infinite, parallel conducting plates, spaced a distance d apart. The region between the plates is filled with a dielectric ε .  Say a voltage V0 is placed across these plates. 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Q: What electric potential field V Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev, electric field E Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev and charge density ρs Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRevis 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.

Q: Yikes! Where do we even start ?
 A: 
We might start with the electric potential field VBoundary Value Problems(Part- 1) - 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- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

b) It must also satisfy the boundary conditions: 

V z =−d =V              V z = 0 = 0

Consider first the dielectric region ( −d <z< 0 ).  

Since the region is a dielectric, there is no free charge, and: 

ρvBoundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev =0

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

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

This problem is greatly simplified, as it is evident that the solution V Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev is independent of coordinates x and y .  In other words, the electric potential field will be a function of coordinate z only: V Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev =V (z )

This make the problem much easier!  Laplace’s equation becomes: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Integrating both sides of Laplace’s equation, we get: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

And integrating again we find: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

We find that the equation V (z ) =C1z + C2 will satisfy Laplace’s equation  (try it!).  We must now apply the boundary conditions to determine the value of constants Cand C2.

We know that the value of the electrostatic potential at every point on the top (z =-d) plate is  V (-d)=V0, while the electric potential on the bottom plate (z =0) is zero (V (0) =0 ).  Therefore: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Two equations and two unknowns (C1 and C2)! Solving for C1 and C2 we get: 

Boundary Value Problems(Part- 1) - 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- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Before we proceed, let’s do a sanity check!

In other words, let’s evaluate our answer at z = 0 and z = -d, to make sure our result is correct: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

and 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Now, we can find the electric field within the dielectric by taking the gradient of our result: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

And thus we can easily determine the electric flux density by multiplying by the dielectric of the material: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

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

Q: Charge density !?!  I thought that we already determined that the charge density ρvBoundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev is equal to zero?

A: We know that the free charge density within the dielectric is zero—but there must be charge somewhere, otherwise there would be no fields! 

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- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

First, we find that the electric flux density on the bottom surface of the top conductor (i.e., at z = −d ) is: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

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

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

 Therefore, we find that the surface charge density on the bottom surface of the top conductor is: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

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

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Therefore, evaluating the electric flux density on the top surface of the bottom conductor (i.e., z = 0 ), we find: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

We should note several things about these solutions: 

1. Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

2. Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

3. D Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRevandE Boundary Value Problems(Part- 1) - 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 using superposition and eq. 4.20 in section 4-5! 

I.E.: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

 

Example: Charge Filled Parallel Plates

Consider now a problem similar to the previous example (i.e., dielectric filled parallel plates), with the exception that the space between the infinite, conducting parallel plates is filled with free charge, with a density: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Q: How do we determine the fields within the parallel plates for this problem?

A: Same as before!  However, since the charge density between the plates is not equal to zero, we recognize that the electric potential field must satisfy Poisson’s equation: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

For the specific charge density Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Since both the charge density and the plate geometry are independent of coordinates x and y, we know the electric potential field will be a function of coordinate z only (i.e., V Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev =V (z) ).

Therefore, Poisson’s equation becomes: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

We can solve this differential equation by first integrating both sides: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

And then integrating a second time: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Note that this expression for VBoundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev satisfies Poisson’s equation for this case.  The question remains, however: what are the values of constants C1 and C?

We find them in the same manner as before—boundary conditions!

Note the boundary conditions for this problem are: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Therefore, we can construct two equations with two unknowns: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

It is evident that C2 = 0, therefore constant C1 is: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

The electric potential field between the two plates is therefore: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

Performing our sanity check, we find: 

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

and  

Boundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev

From this result, we can determine the electric field EBoundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev, the electric flux density DBoundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev, and the surface charge density ρ sBoundary Value Problems(Part- 1) - Electrostatics, Electromagnetic Theory, CSIR-NET Physical Science Physics Notes | EduRev, as before.
Note, however, that the permittivity of the material between the plates is ε0, as the “dielectric” between the plates is freespace. 

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