The Net Electric Field Inside a Dielectric

AI Thread Summary
The discussion revolves around the net electric field inside a dielectric and the surface charge density on its outer surface. The net electric field is expressed as a function of the applied electric field and the polarization vector, leading to a relationship between the charges on the inner and outer surfaces. Participants debate the correct form of the surface charge density, with one asserting that the correct answer includes the dielectric constant in the denominator, while another argues that it should not. The conversation highlights the transition of dielectric behavior to that of a conductor as the dielectric constant approaches infinity, influencing the surface charge density calculation. Ultimately, the participants seek clarity on the derivation and implications of their calculations regarding the electric field and surface charge density in different dielectric scenarios.
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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
$$\vec E_{net} = \frac{\vec E_{applied}}{\varepsilon_r}$$
1635748989753.png

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

what do you mean by
anuttarasammyak said:
If dielectric surface is metal dielectric constant is + infinity and we would observe opposite sign charge so I choose (d).
 
I regard conductor has infinite dielectric constant. Please imagine shielding by surrounding conductor.
 
anuttarasammyak said:
Please imagine shielding by surrounding conductor.
1635760877063.png

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.
 
1635762340023.png

So the surface charge density is
\frac{Q}{4\pi r_2^2} for infinite ##\epsilon_r##. (d)
 
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anuttarasammyak said:
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?
 
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##.
 
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anuttarasammyak said:
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
 
  • #10
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.
 
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  • #11
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*}
 
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