# Griffiths page 182

[SOLVED] Griffiths page 182

1. Homework Statement
This question refers to Griffiths E and M book.

This page has been confusing me immensely over the past two days. First, can someone explain the logic in the statement below Figure 4.21 "so D can be found from the free charge as though the dielectric were not there". I do not understand at all the justification for the equation that follows.
Furthermore, what is most confusing to me is when we can assume the curl of D vanishes. Griffiths makes it seem like this is almost never true, but then in all of the problems he uses this fact like in Example 4.6. Can someone explain why he uses this fact in Example 4.6 and also when I can use eqn 4.35 in exercises?

2. Homework Equations

3. The Attempt at a Solution

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Ben Niehoff
Gold Member
Read the paragraph in the middle of the page. $\vec \nabla \times \vec D = 0$ everywhere except on the boundary between two media of different permittivity. On the boundaries between media, the curl of D is a delta function such that the line integral (shown in the picture) satisfies Stokes' theorem.

Note that this assumes that the media are linear and isotropic. If there is a more complicated relationship between E and D, then the curl of D might be nonzero within the medium as well.

Currently I am busy.So I will give my reply this afternoon.

First, can someone explain the logic in the statement below Figure 4.21 "so D can be found from the free charge as though the dielectric were not there". I do not understand at all the justification for the equation that follows.

First realize one thing:Coulomb's law (#Griffiths equation 2.8) is a solution of two differential equations

$$\bigtriangledown\ . \ E=\frac{\rho}{\epslion_0}$$ and $$\bigtriangledown\times\ E=\ 0$$

Supplied with sufficient boundary conditions.

So,the form of solution of those two equations is provided by equation 2.8

Then bring the analogy:

$$\bigtriangledown\ . \ D=\rho_f$$ and $$\bigtriangledown\times\ D=\ 0$$

will give you a solution like Coulomb's law.

The question is what is the form of the "Coulomb's law"?

Griffiths has given that expression in his book.Refer to section 4.3.2.Look at the "inequation" giving the value of D(r).

This is the requisite Coulomb's law for you.You need to replace the inequality sign by equality sign.And if you go through section 4.3.2, you will find this plausible.Because, inside of a linear homogeneous dielectric, curl of D and hence, curl of P is zero.Note that Griffiths indirectly said that if curl of P were zero, you would have that Gauss law for D.

Now how to reconcile this solution with $$\ D=\epsilon_0\ E_{\ vac}$$?Note that the solution D(r) is infact $$\ D=\epsilon_0\ E_{\ vac}$$ itself.

Then equation 4.35 follows easily.

So,this part is clear.Now the following part:

Furthermore, what is most confusing to me is when we can assume the curl of D vanishes. Griffiths makes it seem like this is almost never true, but then in all of the problems he uses this fact like in Example 4.6. Can someone explain why he uses this fact in Example 4.6 and also when I can use eqn 4.35 in exercises?
Curl D vanishes when curl P vanishes.And this happens when you have the planar,cylindrical or spherical symmetries---as mentioned in last paragraph of 4.3.2.

Why?Because, in that case the ELECTRIC FIELD pattern is such that curl of electric field lines vanishes.Imagine the field lines for infinite parallel sheet (example 4.6) or infinitely long cylindrical charge distribution or spherical...
In each of them,you get curl D=0.So, no problem in using that equation.In these cases, you can use equation 4.35.

You CANNOT use 4.35 or curl D=0 where there is a fringing field due to absence of symmetry.Very common example is finite cylindrical charge distribution due to absence of
symmetry.refer to problem 4.11 and try to draw the diagrams.You will see that the field lines do have non-zero curl and then, you cannot use those equation...and the Gauss law for D again becomes an inequation as told by Griffiths.This is the general case.

I hope the problem is now clear to you.

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Sorry. I still do not see how Griffith got this:

"so D can be found from the free charge as though the dielectric were not there:

$$\bf D = \epsilon_e E_{vac}$$
"

Can someone please help me with this?!? This has been bothering me for such a long time!!!

someone please! you have no idea how much this bothers me!

anyone?

anyone?

Mr. ehrenfest,

I had given you the correct answer.In fact by answering you,I clarified the matter to myself.I have done a course on EM using Griffiths two years ago.I know that sometimes Griffiths says something that is not always easy to understand.But what he says is CORRECT almost everywhere.You have to try YOURSELF.

If you tried to understand what I wrote,you could see the explanation.But you did not exerted yourself enough to understand.

someone please! you have no idea how much this bothers me!
Understand that other people have work---they are not here to do everything for you.

First realize one thing:Coulomb's law (#Griffiths equation 2.8) is a solution of two differential equations

$$\bigtriangledown\ . \ E=\frac{\rho}{\epslion_0}$$ and $$\bigtriangledown\times\ E=\ 0$$

Supplied with sufficient boundary conditions.

So,the form of solution of those two equations is provided by equation 2.8

Then bring the analogy:

$$\bigtriangledown\ . \ D=\rho_f$$ and $$\bigtriangledown\times\ D=\ 0$$

will give you a solution like Coulomb's law.

The question is what is the form of the "Coulomb's law"?

Griffiths has given that expression in his book.Refer to section 4.3.2.Look at the "inequation" giving the value of D(r).

This is the requisite Coulomb's law for you.You need to replace the inequality sign by equality sign.And if you go through section 4.3.2, you will find this plausible.Because, inside of a linear homogeneous dielectric, curl of D and hence, curl of P is zero.Note that Griffiths indirectly said that if curl of P were zero, you would have that Gauss law for D.

Now how to reconcile this solution with $$\ D=\epsilon_0\ E_{\ vac}$$?Note that the solution D(r) is infact $$\ D=\epsilon_0\ E_{\ vac}$$ itself.

Then equation 4.35 follows easily.

So,this part is clear.Now the following part:
Thanks neelakash, I finally see what is going on. I was struggling with this for about two weeks! I whats above several times and it was almost totally incomprehensible to me until 5 minutes ago! Wow. Let me rephrase what you wrote in my own words to see if I really understand. You gave basically two reasons why eqn 4.35 is true:

1) Solutions to Laplace's equation and Poisson's equations are unique if you impose boundary conditions. And we know that E_{vac} must solve the equations

$$\bigtriangledown\ . \mathbf{ v} \epsilon_0=\rho_f$$ and $$\bigtriangledown\times\ \mathbf{ v} \epsilon_0=\ 0$$

Therefore,
$$\mathbf{D} = \epsilon E_{vac}$$

Very nice.

2) On page 178, Griffiths says "For the divergence alone is not sufficient to determine a vector field; you need to know the curl as well". And now we do know the curl! So everything makes sense. Thanks again neelakash for pushing to me to think harder in the post above this one.

Yes...the idea is that the familiar laws like---Coulomb's law and Biot-Svavart's laws are

basically the SOLUTIONS of the fundamnetal vetor-differential equations (given boundary conditions).

You have the differential equations for the vector D(r).One is divergence and another is curl.If you have the boundary conditions in hand, you can have the Coulomb's law for D what Griffiths write in page 178

If curl of D is non zero (which is the general case), you have (p178)

$$\ D(r) \neq\frac{1}{\ 4 \pi}\int\frac{\hat{R}}{\ R^2 }{\rho_f}(\ r' )\ dV$$

And if curl of D (as well as curl of P) equal to zero (in a special case), you have

$$\ D(r) = \frac{1}{\ 4 \pi}\int\frac{\hat{R}}{\ R^2 }{\rho_f}(\ r' )\ dV$$

Note that these Coulomb's Laws for D do not have $$\epsilon_0$$.

If curl of D (as well as curl of P) equal to zero (as in your problem)

You CAN use the 2nd formula.

Note also that the second formula is exactly $$\ D=\epsilon_0\ E_{\ vac}$$

Because the charge density $$\rho_f$$ (NOT $$\rho$$) would have produced that much vacuum electric field.

So it's done!

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Exert yourself