A problem in Griffith's textbook of electrodynamics, P189.

[karlzr]

Homework Statement

Exercise 1:
Suppose a infinite plane z=0 divides the 3-dimensional space into two parts. The region below z=0 is filled with linear dielectric material of susceptibility $\chi$. The bound charge density at z=0 is $\sigma$. Find the electric field of the two regions.

Exercise 2:
The original problem in Griffith's book is to calculate the force on a point charge q situated a distance d above the the origion (x=0,y=0,z=d). The first step is to find the electric field $E_z$ just inside the dielectric at z=0, which is due in part to q and in part to the bound charge itself. And he states that the latter contribution is
$$-\sigma/2\epsilon_0$$.


In my opinion, we should write two equations(from Guass's law in terms of E and D respectively):
$$\epsilon_0 E_1+\epsilon_0(1+\chi)E_2=0$$
$$\epsilon_0(E_1+E_2)=\sigma$$
Then $E_2$ is my answer, which is quite different from the expression in Griffith's book $-\sigma/2\epsilon_0$.

I need a detailed solution to the electric field of this configuration.

 

[Ethan0718]

What's the meaning of those two equations you write?

 

[gabbagabbahey]

In my opinion, we should write two equations:
$$\epsilon_0 E_1+\epsilon_0(1+\chi)E_2=0$$
$$\epsilon_0(E_1+E_2)=\sigma$$
Then $E_2$ is my answer, which is quite different from the expression in Griffith's book $-\sigma/2\epsilon_0$.

Why, in your opinion, should we write these two equations?

Keep in mind that the boundary condition Eq 4.40 applies to the normal component of the total electric field when crossing a surface from one linear dielectric to another, not just the part of the (normal component) field due only to the surface charge there (which is what Griffiths' claims is $\frac{\sigma}{2\epsilon_0}$).

 

[karlzr]

What's the meaning of those two equations you write?

These two equations are two expressions of Gauss's law in terms of E and D respectively.

 

[karlzr]

Why, in your opinion, should we write these two equations?

Keep in mind that the boundary condition Eq 4.40 applies to the normal component of the total electric field when crossing a surface from one linear dielectric to another, not just the part of the (normal component) field due only to the surface charge there (which is what Griffiths' claims is $\frac{\sigma}{2\epsilon_0}$).

I modified my question as you can see. Now as to Exercise 1, do these two equtions hold?
Then the total electric field $E_z$ should be the summation of the two parts.

 

[Ethan0718]

You're right! However, The "E2" you write is not the contribution of the induced charge. The eletric field contribution of the induced charge is - sigma/(2*epsilon)

The E2 which you write is the "total" electric field just below the plate due to charge q and those induced charge.

 

[karlzr]

You're right! However, The "E2" you write is not the contribution of the induced charge. The eletric field contribution of the induced charge is - sigma/(2*epsilon)

The E2 which you write is the "total" electric field just below the plate due to charge q and those induced charge.

I changed the way I put my question. What I don't understand is why the contribution from the bound charge is $-\sigma/(2\epsilon_0)$. That is why I put forward Exercise 1, to consider the part of electric field from the bound charge. In this case, I think the equations are OK. Then $E_2$ is the result we are looking for, which can be added directly to the total electric field because of superposition principle.