1
$\begingroup$

This is problem 4.25 from Griffiths' electrodynamics, it says:

Suppose the region above the $xy$ plane is filled with linear dielectric of susceptibility $\chi_e{}'$ , and the region below the $xy$ plane is filled with linear dielectric of susceptibility $\chi_e$ and a point charge $q$ is situated a distance $d$ above the origin.

enter image description here

Now I know that due to $q$ there will be bound surface charge densities induced on the upper surface of the lower dielectric and on the lower surface of the upper dielectric.

And to get the bound charge on the lower dielectric, $ \sigma_b = \vec P . \hat n = \epsilon_0 \chi_e E_z$. While $\sigma_b' = ?$

I understand how to get $\sigma_b$ as the following:

$$\sigma_b = \epsilon_0 \chi_e \left[-\dfrac{1}{4 \pi \epsilon_0}\dfrac{qd/ \epsilon_r'}{(r^2 +d^2)^{\frac{3}{2}}}-\dfrac {\sigma_b}{2 \epsilon_0} -\frac{ \sigma_b'}{2 \epsilon_0}\right]$$

Inside these brackets are the electric field due to $q$ which points in ( $ - \hat z $) and the electric field due to $\sigma_b$ which also points in ( $ -\hat z$)

But I have a problem figuring out $ \sigma_b'$

$\sigma_b' = \vec P \cdot \hat n $

but what is the direction of polarization and how is $ \hat n$ oriented? The solution to $\sigma_b'$ is:

$$\sigma_b' = \epsilon_0 \chi_e' \left[ \dfrac{1}{4 \pi \epsilon_0}\frac{qd/ \epsilon_r'}{(r^2 +d^2)^{\frac{3}{2}}} - \dfrac { \sigma_b}{2 \epsilon_0} -\frac{ \sigma_b'}{2 \epsilon_0}\right]$$

But I don't understand how to get the signs right in that last bracket, I have trouble dealing with the upper dielectric, how is its normal vector oriented and what is the direction of polarization or what exactly are the directions of the electric fields? Hopefully you can help me.

$\endgroup$
  • $\begingroup$ Note to answerers: acceptable answers to homework-like questions focus on the conceptual parts of the question. Complete solutions to homework problems may be tempoarily removed; our community is not interested in becoming a homework help service. $\endgroup$ – rob Jul 16 '19 at 4:00
  • $\begingroup$ Can you recover the answer please? I was having confusion with how the normal vectors were oriented and the concept of induced surface charge. $\endgroup$ – khaled014z Jul 16 '19 at 4:29
  • $\begingroup$ @rob To be honest, it was more like an illustrated example. I understand the policy against being a homework-solving site, but some concepts are best explained through an example. Anyway, I'll dig into meta.phys.SE to find if this was argued about before. $\endgroup$ – acarturk Jul 16 '19 at 8:39
1
$\begingroup$

The formula $\sigma_b = \vec{P}·\hat{n}$ assumes that $\hat{n}$ is the outward normal vector, i.e. it is pointing away from the dielectric you are calculating the bound surface charge density for.

I believe you implicitly made use of this when writing the expression for $\sigma_b$, for which $\hat{n} = \hat{z}$.

For the upper dielectric, $\hat{n} = -\hat{z}$. You can verify that this gives the correct signs in the expression for $\sigma_b'$.

$\endgroup$
  • $\begingroup$ I got the normal vectors right, but now I'm having trouble with the direction of the electric fields, there should be positive surface charge on the lower surface of the upper dielectric and negative surface charge on the upper surface of the lower dielectric, and their fields don't add up outside as they should, am I misunderstanding something? $\endgroup$ – khaled014z Jul 16 '19 at 16:16
  • $\begingroup$ I realized the sign of the first term in your expression for $\sigma_b'$ is incorrect, as you noticed. In the upper dielectric, the E-field due to $q$ has a positive $z$ component, therefore $\vec{E}_q ·\hat{n}$ is negative, since $\hat{n} = -\hat{z}$. On an unrelated note, I don't know what $\epsilon_r'$ means but it should be $1$, since you are adding E-fields due to all charges, free and bound. The susceptibilities in your expressions already take dielectric constants into account. $\endgroup$ – Puk Jul 16 '19 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.