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Image I'm referring to:

![enter image description here

In the figure both shells are conductors.

Let's say I have to find the electric field at a point P as shown in the figure. I thought, since this point lies inside a conductor, the Electric field at this point has to be $0$ no matter what. But I was taught that there will be no electric field at P due to the outer shell but there will be a electric field at point P due to the inner shell, (because it has some charge).

I'm confused now, beacuse earlier I was told the electric field inside a conductor is $0$, and that point clearly lies inside the conductor, where am I going wrong? Won't there be induction of $-Q_{1}$ on the inner surface of the outer shell, in such a way that the electric field inside the conductor becomes $0$?

Also, by the term inside the conductor mean in the bulk or in the material of the conductor or something else?

Please help me out, I have studied from many sources, any no one cares to be critical about their words, hence the confusion.

Edit: For all future readers, this image is a good explaination of what is going on inside such situations. ![enter image description here

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    $\begingroup$ My suggestion to reduce the confusion is to rely on the most important law that governs these systems, forgetting about other statements that are sometimes correct and sometimes wrong. I am talking about Gauss' law, which states that the flux of electric field through a closed surface is equal to the charge inside that surface divided by $\varepsilon_0$. $\endgroup$
    – Matteo
    Commented Dec 9, 2021 at 11:27
  • $\begingroup$ @Matteo If I use Gauss' law here, yes, there actaully is an Electric Field at point P. But still it would be useful if someone could explain what exactly happens in situations like these or in cases like a cavity inside a solid conductor. $\endgroup$ Commented Dec 9, 2021 at 11:32
  • $\begingroup$ Ok, I'll try to elaborate an answer then :) $\endgroup$
    – Matteo
    Commented Dec 9, 2021 at 11:35
  • $\begingroup$ @Matteo Thank you very much., Also, I read more and finally starting to understand some stuff. Actually I now realise that by inside the conductor, we mean, the bulk or the material of the conductor. And say if we have a cavity with some charge inside a conductor, the cavity will have some electric field inside it but the bulk of the conductor will not have any electric field. More: youtube.com/watch?v=3nITB_C9lgg. $\endgroup$ Commented Dec 9, 2021 at 11:43
  • $\begingroup$ Point $P$ does not lie inside of a conductor - we have two thin spherical shells here, but the space between the shells is vacuum/air, not a conductor. Your textbook surely explain why the field inside the conductor is zero, and knowing this explanation immediately answers your question of what inside the conductor actually means. $\endgroup$
    – Roger V.
    Commented Dec 9, 2021 at 11:49

2 Answers 2

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The most reliable way to approach these problems is to use fundamental laws of electromagnetism, in particular Gauss' law and linearity of the electric field. A thorough application of these laws can answer all of your questions.

Gauss law

The electric field flux through a closed surface $\Phi_E$ is given by the charge inside that surface $Q_{int}$ divided by $\varepsilon_0$: $$\Phi_E = \frac{Q_{int}}{\varepsilon_0 }. $$

Linearity of electric field

The total electric field at a point is the vector sum of all the electric field produced by the single sources. If there are two sources labeled $1$ and $2$, the field at a point $\vec{E}(P)$ is

$$ \vec{E}(P) = \vec{E}_1(P) + \vec{E}_2(P). $$

Let's now use these tools to solve your problem: at point P there are in principle two contributions to the field, one coming from the inner shell, and one coming from the outer shell. The former has modulus $Q_1/4\pi\varepsilon_0r^2$ and the latter vanishes, as a consequence of Gauss' law (exercise). Bottom line: there is an electric field at P, only coming from the inner shell charges.

Now let's comment your statements one by one:

I thought, since this point lies inside a conductor, the Electric field at this point has to be 0 no matter what.

This is a misunderstanding, since "inside a conductor" means "in the bulk" of a conductor, but here there's no bulk at all.

I was taught that there will be no electric field at P due to the outer shell but there will be a electric field at point P due to the inner shell

This is correct, as discussed above.

EDIT while writing this answer, I was unaware of most of the comments and other answers, so there might be some redundancy.

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  • $\begingroup$ no problem, everything is clear now. I will cut down on some content tomorrow so that question is more focused. Thank you for your help. $\endgroup$ Commented Dec 9, 2021 at 11:59
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As Matteo points out, it is better to use Gauss's law to avoid any confusion.

...earlier I was told the electric field inside a conductor is $0$

This is true only when put in conjunction with another statement you may have come across:

If the conductor is isolated and carries a charge, the charge resides on it's surface.

Hence, if we apply Gauss's law to a gaussian sphere inside the conductor, as $q_\text{in} = 0$ it follows that $E = 0$ ($\vec E$ and $d\vec A$ are not perpendicular)

Now, in the question, if you draw a gaussian sphere that is concentric with either shell and passes through P, you will find that $q_\text{in} = Q_1$ is not zero, and hence, $|E| > 0$

Hope this helps.

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  • $\begingroup$ Thank you as well, also you mention that you hate chemistry, same here. That thing makes me want to quit studying. Maths is the best!! $\endgroup$ Commented Dec 9, 2021 at 12:11
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    $\begingroup$ Glad to know that I'm not alone :-) $\endgroup$
    – Cross
    Commented Dec 9, 2021 at 12:40
  • $\begingroup$ You cannot use Gauss' law to show that the field inside the conductor is zero. Gauss's law only tells that the flux through the closed surface is zero. This is all. For the situation shown in the OP, with the thick conductive shell, drawing the gaussian is missleading. The field is zero due to the electroostatic equilibrium condition in a conductive medium. $\endgroup$
    – nasu
    Commented Dec 9, 2021 at 13:26

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