Undergrad Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)

[Plantation]

TL;DR

Understanding the Griffith, Electrodynamics, Example 4.8.

I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused.

> Example 4.8. Suppose the entire region below the plane $z=0$ in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility $\chi_e$. Calculate the force on a point charge $q$ situated a distance $d$ above the origin.

Solution : The surface bound charge on the $xy$ plane is of opposite sign to $q$, so the force will be attractive. ( In view of Eq.4.39, there is no volume bound charge.)

Q.1. Can anyone explain why this sentence is true more friendly?

( Cont. ) Let us first calculate $\sigma_b$, using Eqs. 4.11 and 4.30.
$$ \sigma_b = \mathbf{P} \cdot \hat{\mathbf{n}} = P_z = \epsilon_0 \chi_e E_z,$$
where $E_z$ is the $z$-component of the total field just inside the dielectric, at $z=0$.
This field is due in part to $q$ and in part to the bound charge itself ( Q.2. Why? )

( Cont. ) From Coulomb's law, the former contribution is

$$ -\frac{1}{4 \pi \epsilon_0}\frac{q}{(r^2 + d^2)} \cos \theta = -\frac{1}{4 \pi \epsilon_0}\frac{qd}{(r^2+d^2)^{3/2}}$$

, where $r= \sqrt{x^2+y^2}$ is the distance from the origin.

Q.3. Why the former contribution is given by such that? Why the cosine term is appeared?

And through some argument, he deduced that

$$ \sigma_b = -\frac{1}{2 \pi} (\frac{\chi_e}{\chi_e + 2}) \frac{qd}{(r^2+d^2)^{3/2}}. \tag{4.50} $$

( Cont. ) Apart from the factor $\chi_e / ( \chi_e + 2)$, this is exactly the same as the induced charge on an infinite conducting plane under similar circumstances ( Eq. 3.10 ). Evidently the total bound charge is
$$ q_b = -(\frac{\chi_e}{\chi_e +2 })q \tag{4.51}$$

Q.4. Why the total bound charge is given by $(4.51)$ ? What is definition of total bound charge?

Can anyone teach me?

 

[kuruman]

A.1 The surface bound charge on the $xy$ plane is of opposite sign to $q$, so the force will be attractive.
Unlike a conductor, the external electric field is non-zero inside the dielectric. The atoms making up the dielectric are polarized and produce an opposing field that is not large enough to cancel the external field (see figure, not mine, on the right). There is no net charge on the dielectric, but the separation of positive and negative charges (polarization) on the surface will result in an attractive force on a positive charge placed to the left of the dielectric. That's because the negative charges are closer to the external charge than the positive charges.

A.2 This field is due in part to $q$ and in part to the bound charge itself.
Look at the drawing. The net field inside is the vector sum of the external field and the field generated by the polarized atoms.

A.3. Why the former contribution is given by such that? Why the cosine term is appeared?
It is the vertical component of the electric field exerted by a dielectric surface element on the charge. That is why the cosine. The horizontal component of the Coulomb force is cancelled by the symmetrically-located surface element on the other side. The net horizontal force on the charge is zero by symmetry.

A.4 Why the total bound charge is given by $(4.51)$? What is definition of total bound charge?
The total bound charge can be found by integrating the surface density over the entire area of the slab.

Using polar coordinates, $$q_b=\int \sigma_b~dA= -\frac{1}{2 \pi} (\frac{\chi_e}{\chi_e + 2})qd \int_0^{\infty}\frac{1}{(r^2+d^2)^{3/2}}2\pi ~ r~dr.$$

 

[TSny]

@kuruman has addressed your questions. But I’ll add a few additional remarks.

Solution : The surface bound charge on the $xy$ plane is of opposite sign to $q$, so the force will be attractive. ( In view of Eq.4.39, there is no volume bound charge.)

Q.1. Can anyone explain why this sentence is true more friendly?

There are two sentences in your quote. Are you asking why the force is attractive, or are you asking why there is no volume bound charge?

( Cont. ) Let us first calculate $\sigma_b$, using Eqs. 4.11 and 4.30.
$$ \sigma_b = \mathbf{P} \cdot \hat{\mathbf{n}} = P_z = \epsilon_0 \chi_e E_z,$$
where $E_z$ is the $z$-component of the total field just inside the dielectric, at $z=0$.
This field is due in part to $q$ and in part to the bound charge itself ( Q.2. Why? )

The dielectric extends throughout the entire region below the plane $z = 0$. Also, there isn’t any volume bound charge at any point within the dielectric. So, the only charges are the bound, surface charge on the dielectric at $z = 0$ and the point charge $q$. So, the total field at any point inside or outside the dielectric will be the superposition of the field from the surface bound charge and the field from $q$.

( Cont. ) From Coulomb's law, the former contribution is

$$ -\frac{1}{4 \pi \epsilon_0}\frac{q}{(r^2 + d^2)} \cos \theta = -\frac{1}{4 \pi \epsilon_0}\frac{qd}{(r^2+d^2)^{3/2}}$$

, where $r= \sqrt{x^2+y^2}$ is the distance from the origin.

Q.3. Why the former contribution is given by such that? Why the cosine term is appeared?

Griffiths is considering the z-component of the electric field due to the point charge $q$ at a point just inside the dielectric:

The factor of $\cos \theta$ comes from finding the z-component of the electric field $\mathbf E$.

And through some argument, he deduced that

$$ \sigma_b = -\frac{1}{2 \pi} (\frac{\chi_e}{\chi_e + 2}) \frac{qd}{(r^2+d^2)^{3/2}}. \tag{4.50} $$

( Cont. ) Apart from the factor $\chi_e / ( \chi_e + 2)$, this is exactly the same as the induced charge on an infinite conducting plane under similar circumstances ( Eq. 3.10 ). Evidently the total bound charge is
$$ q_b = -(\frac{\chi_e}{\chi_e +2 })q \tag{4.51}$$

Q.4. Why the total bound charge is given by $(4.51)$ ? What is definition of total bound charge?

The definition of total bound charge, $q_b$, is the sum total of all of the bound charge on the infinite plane $z = 0$.

You could find $q_b$ by setting up the integral of $\sigma_b$ over the $z = 0$ plane, as explained by @kuruman. However, I believe that Griffiths wants you to see that (4.51) comes directly from (4.50) and equations (3.10) and (3.11) from chapter 3. This doesn’t require you to do any integration. In (3.11), Griffiths already did an integral very similar to @kuruman's.

So far in Griffiths' solution, he has not yet attempted to find the force on the point charge, which is what the problem statement asks for. He states below equation (4.51) that you could use equation (4.50) for $\sigma_b$ to set up an integral to find the electric field $\mathbf E$ at the location of $q$ due to all of the bound charge. The force on the point charge is then $\mathbf F = q \mathbf E$. However, he does not carry out this approach to finding $\mathbf F$.

Instead, he outlines how to find $\mathbf F$ using the method of images. Here he uses the result (4.50) for $\sigma_b$.

 

[Plantation]

@kuruman has addressed your questions. But I’ll add a few additional remarks.

There are two sentences in your quote. Are you asking why the force is attractive, or are you asking why there is no volume bound charge?


The dielectric extends throughout the entire region below the plane $z = 0$. Also, there isn’t any volume bound charge at any point within the dielectric. So, the only charges are the bound, surface charge on the dielectric at $z = 0$ and the point charge $q$. So, the total field at any point inside or outside the dielectric will be the superposition of the field from the surface bound charge and the field from $q$.


Griffiths is considering the z-component of the electric field due to the point charge $q$ at a point just inside the dielectric:

View attachment 366380

The factor of $\cos \theta$ comes from finding the z-component of the electric field $\mathbf E$.


The definition of total bound charge, $q_b$, is the sum total of all of the bound charge on the infinite plane $z = 0$.

You could find $q_b$ by setting up the integral of $\sigma_b$ over the $z = 0$ plane, as explained by @kuruman. However, I believe that Griffiths wants you to see that (4.51) comes directly from (4.50) and equations (3.10) and (3.11) from chapter 3. This doesn’t require you to do any integration. In (3.11), Griffiths already did an integral very similar to @kuruman's.

So far in Griffiths' solution, he has not yet attempted to find the force on the point charge, which is what the problem statement asks for. He states below equation (4.51) that you could use equation (4.50) for $\sigma_b$ to set up an integral to find the electric field $\mathbf E$ at the location of $q$ due to all of the bound charge. The force on the point charge is then $\mathbf F = q \mathbf E$. However, he does not carry out this approach to finding $\mathbf F$.

Instead, he outlines how to find $\mathbf F$ using the method of images. Here he uses the result (4.50) for $\sigma_b$.

For the question 1, Both :)

 

[TSny]

For the question 1, Both :)

Intuitively, the reason that $\sigma_b$ is negative is because the field of the positive, external charge $q$ tends to pull negative charge in the dielectric molecules toward $q$. So, the negative end of the polarized molecules "poke out" at the surface of the dielectric.

It's really more complicated, since the actual field that polarizes a molecule is not just the field of $q$. The field that polarizes a molecule is the superposition of the field of $q$ and the field produced by all of the other polarized molecules! This is discussed in section (4.2.3), which is not an easy section.

But note that Griffiths' derivation of (4.50) for $\sigma_b$ does not rely on any assumption regarding the sign of $\sigma_b$. So, the negative result for $\sigma_b$ given by (4.50) demonstrates that $\sigma_b$ is, in fact, negative.

Your other question is why the bound, volume charge density must be zero. Griffiths says that this is due to (4.39) (which is valid for a linear dielectric): $$\rho_b = -\nabla \cdot \mathbf P = -\nabla \cdot\left(\epsilon_o \frac{\chi_e}{\epsilon} \mathbf D \right) = -\left(\frac{\chi_e}{1+\chi_e}\right)\rho_f.$$
Is there a particular equality here that you don't understand? Note that $\rho_f = 0$ by assumption in example 4.8.

 

[Plantation]

Intuitively, the reason that $\sigma_b$ is negative is because the field of the positive, external charge $q$ tends to pull negative charge in the dielectric molecules toward $q$. So, the negative end of the polarized molecules "poke out" at the surface of the dielectric.

It's really more complicated, since the actual field that polarizes a molecule is not just the field of $q$. The field that polarizes a molecule is the superposition of the field of $q$ and the field produced by all of the other polarized molecules! This is discussed in section (4.2.3), which is not an easy section.

But note that Griffiths' derivation of (4.50) for $\sigma_b$ does not rely on any assumption regarding the sign of $\sigma_b$. So, the negative result for $\sigma_b$ given by (4.50) demonstrates that $\sigma_b$ is, in fact, negative.

Your other question is why the bound, volume charge density must be zero. Griffiths says that this is due to (4.39) (which is valid for a linear dielectric): $$\rho_b = -\nabla \cdot \mathbf P = -\nabla \cdot\left(\epsilon_o \frac{\chi_e}{\epsilon} \mathbf D \right) = -\left(\frac{\chi_e}{1+\chi_e}\right)\rho_f.$$
Is there a particular equality here that you don't understand? Note that $\rho_f = 0$ by assumption in example 4.8.

Thank you ! For the final sentence in your answer, why $\rho_f = 0$ is by assumption in example 4.8? Is it because filling the entire region below the plane $z=0$ is 'dielectric' material?

 

[TSny]

Thank you ! For the final sentence in your answer, why $\rho_f = 0$ is by assumption in example 4.8? Is it because filling the entire region below the plane $z=0$ is 'dielectric' material?

That's not the reason. The assumption is more general.

Free charge inside the volume of a dielectric material would have to be "embedded" there by some means. So, in textbook problems, we assume no free charge inside dielectrics unless explicitly stated otherwise. Note the comment in the textbook just below equation (4.39).