The Net Electric Field Inside a Dielectric

[Hamiltonian]

Homework Statement

A conducting sphere of radius r1 is surrounded by a dielectric layer of outer radius r2 and dielectric constant e_r if the conducting sphere is given a charge q, determine the surface charge density of polarization charges on the outer surface of the dielectric layer.

Relevant Equations

$$\vec E_{net} = \frac{\vec E_{applied}}{\varepsilon_r}$$

The net Electric field(inside the dielectric):
$$E_{net} = \frac{1}{4\pi \varepsilon_0 \varepsilon_r} \frac{q}{r^2}$$
$$\vec E_{net} = \vec E_{applied} - \vec p$$
where p is the polarization vector.
let charge $q_{-}$ be present on the inner surface of dielectric and $q_{+}$ on the outer surface of the dielectric.
$$\vec p = \frac{q_{-}}{4\pi \varepsilon_0 \varepsilon_r r^2}$$
$$\frac{1}{4\pi \varepsilon_0 \varepsilon_r} \frac{q}{r^2} = \frac{1}{4\pi \varepsilon_0}\frac{q}{r^2} - \frac{q_{-}}{4\pi \varepsilon_0 \varepsilon_r r^2}$$
after simplifying you get:
$$q_{-} = q(\varepsilon_r - 1)$$
$$|q_{-}| = |q_{+}|$$
therefore the final answer(surface charge density on the the outer surface of the dielectric):
$$\sigma_{+} = \frac{q_{+}}{4\pi{ r_2}^2} = \frac{(\varepsilon_r - 1)q}{4\pi {r_2}^2}$$
I have obtained option (c) but the correct answer is (d), I am unable to see why there should be $\varepsilon_r$ in the denominator of my final answer, is there a mistake in the answer given?
or have I severely messed up some where?

 

[anuttarasammyak]

If dielectric surface is vacuum dielectric constant is 1 and we would observe no charge so I think (c) or (d) is a right answer.
If dielectric surface is metal dielectric constant is + infinity and we would observe opposite sign charge so I choose (d).

 

[Hamiltonian]

I am unable to understand what you are trying to say.

what do you mean by

If dielectric surface is metal dielectric constant is + infinity and we would observe opposite sign charge so I choose (d).

 

[anuttarasammyak]

I regard conductor has infinite dielectric constant.　Please imagine shielding by surrounding conductor.

 

[Hamiltonian]

Please imagine shielding by surrounding conductor.

this is what the setup looks like, the conductor is surrounded by the dielectric and not vice versa so again I am unable to understand what you are trying to say, could you maybe tell me if there is a mistake in the steps of #1.

 

[anuttarasammyak]

So the surface charge density is
$$\frac{Q}{4\pi r_2^2}$$ for infinite $\epsilon_r$. (d)

 

[Hamiltonian]

View attachment 291535
So the surface charge density is
$$\frac{Q}{4\pi r_2^2}$$ for infinite $\epsilon_r$. (d)

no I don't think the charge on the outermost surface will be Q since there hasn't been a dielectric breakdown, the charge density on the outer surface must be $<\frac{Q}{4\pi {r_2}^2}$

edit: ok I think I understood your logic supporting why option (d) is correct but how could we arrive at that from first principles?

 

[anuttarasammyak]

For finite $\epsilon_r$ yes. You see (d) satisfies your inequality. (c) does not.

And please remind that I have said about conductor infinite $\epsilon_r$.

 

[Hamiltonian]

For finite $\epsilon_r$ yes. You see (d) satisfies your inequality. (c) does not.

And please remind that I have said about conductor infinite $\epsilon_r$.

I understood how you are motivating option (d), as $\varepsilon_r \rightarrow \infty$ i.e, dielectric will tend to behave like a conductor and hence the surface charge density will ted to be $\frac{q}{4\pi {r_2}^2}$.
but how could we arrive at (d) what mistake have I made in #1

 

[anuttarasammyak]

Electric field in dielectric material is
$$\frac{1}{4\pi\epsilon}\frac{Q}{r^2}$$
It is superposition of source electric field and minus Polarization. Polarization is
$$\frac{1}{4\pi\epsilon_0}\frac{Q_P}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}-\frac{1}{4\pi\epsilon}\frac{Q}{r^2}$$
where $Q_P$ is surface charge of dielectric. Thus
$$Q_P=(1-\frac{1}{\epsilon_r})Q$$
Please compare it with your thought.

 

[ergospherical]

Another approach. The radial component of $\boldsymbol{D}$ is continuous at the interface $r=r_2$. At $r = r_2 - \delta$ you have $E_{-} = D/(\epsilon_r \epsilon_0)$. At $r = r_2 + \delta$ you have $E_{+} = D / \epsilon_0$. So $E_{+} - E_{-} = (1-\epsilon_r^{-1}) D/\epsilon_0$. But the jump condition on the electric field is $[E] \equiv E_{+} - E_{-} = \sigma / \epsilon_0$, therefore\begin{align*}
\dfrac{\sigma}{\epsilon_0} = \dfrac{D(1- \epsilon_r^{-1})}{\epsilon_0} \implies \sigma = D(1- \epsilon_r^{-1})
\end{align*}By Gauss' theorem, $4\pi r^2 D(r) = q_{\mathrm{f}} \implies D = q / (4\pi r_2^2)$. So\begin{align*}
\sigma = \dfrac{q(1- \epsilon_r^{-1})}{4\pi r_2^2}
\end{align*}