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Let us take an arbitrary conductor having a weird-shaped cavity inside it. Let $+q$ charge be inserted inside the cavity. The field of $+q$ attracts negative charge & repels positive charge; negative charge accumulates on the surface of the cavity while the induced positive charge accumulates on the outer surface of the conductor till the electric field of the $+q$ charge in the cavity is balanced by the field due to the induced charge inside the conductor. Hence, there exists no electric field inside the conductor due to cancellation of the electric field of the charge residing in the cavity & that formed by the induced charges.

But what happens when an external electric field is applied?

Griffiths writes:

No external fields penetrate the conductor; they are canceled at the outer surface by the induced charge there. Similarly, the field due to charges within the cavity is killed off, for all exterior points, by the induced charge on the inner surface.

How can the induced charge on the outer surface cancel the external field? Hasn't the field of the induced charge been used up for balancing the electric field of the charge in the cavity inside the conductor? How can an used up(I mean to say this field has already been balanced by the field of the cavity-charge inside the conductor) electric field be used to balance another field?? I may be wrong in my sense. Please help.

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5 Answers 5

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The field of the induced charges is not used up to cancel the field in the metal of the inner charges, so that it could no longer be used for anything else.

The two effects (field of the charges inside the conductor and the external field) are independent. Each one accumulates charges on the outside of the conductor to kill the field inside of it. The effects are additive, so applying an additional external field simply will displace more charges to the surface of the conductor.

Mathematically speaking, you find the reason for why the effects are additive in the linearity of Maxwells equations. This is also the reason why the principle of superposition works.

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Instead of taking a cavity, take a massive conducting object with the same shape. Now, if there is an external electric field, the charges will rearrange so that the field inside the object is zero and the induced charge will be at the edges of the object.

Now make that object hollow. You've got your cavity again, the induced charge is still at the outer edges and the field inside is still zero.

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  • $\begingroup$ Griffiths writes about electric field of the induced charge separately; the induced negative charge applies $E_{\text{induced}}$ which balances the field of the cavity-charge & the electric field of the positive induced charge $E_{\text{leftover}}$ balances the external field & its magnitude is zero inside the conductor. Why did he consider the field of the induced charge separately & why is the value of$E_{\text{leftover}}$ zero inside the conductor? Can you tell, sir?? $\endgroup$
    – user36790
    Commented Jul 8, 2015 at 9:28
  • $\begingroup$ Why does he consider the field of the induced charge separately? When he talks about the zero electric field inside a conductor in an external electric field, he says that the field between the negative & positive induced charge cancels the external field; here he considers the field due to the induced charge together, then why does he talk of two fields $E_{\text{induced}}$ & $E_{\text{leftover}}$ separately? $\endgroup$
    – user36790
    Commented Jul 8, 2015 at 9:35
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    $\begingroup$ You have $\vec{E}_{original}$ from the original charge(s) (wether inside or outside the cavity, doesn't matter) and $\vec{E}_{induced}$ from the induced charges. Now the total field is $\vec{E}_{original} +\vec{E}_{induced} = \vec{E}_{total}$. This last is what he calls $\vec{E}_{leftover}$. Now, the induced charges are positioned so that $\vec{E}_{total}$ (or $\vec{E}_{leftover}$) is zero inside the conductor once equilibrium is reached. $\endgroup$
    – Dries
    Commented Jul 8, 2015 at 10:16
  • $\begingroup$ He considers the fields separately because doing so allows the problem to be broken into two smaller problems. This is allowed because the fields just add together linearly. $\endgroup$
    – EL_DON
    Commented Sep 4, 2017 at 20:58
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Perhaps a bit of mechanism will help make it understandable.

Image that, for some reason, there is a non-zero field in the conductor. Conductor have the property that they allow charge to move, and that field puts s force on them: so charge will move. It will move along the field lines until it can’t any more. If it hits a surface, it’ll stop there.

In the process, those charges are moving in the direction that reduced the field that’s moving them. You can see this by drawing field lines and charges, or by noticing that they’re moving to reduce the $E^2$ energy in the field.

Eventually, enough charge has moved to zero out the field, and the process stops with no field remaining in the conductor.

It happens very fast, but if the field is changing really fast, I.e. RF, it’s not fully effective before the field changes again. That leads to interesting phenomena too ...

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There are three electric fields interacting with each other:

  1. the field of the positive charge inside the cavity
  2. the field of the negative induced charge on the inner surface
  3. the field of the positive induced charge on the outer surface.

The first two will cancel each other on all the exterior points of the inner surface while the latter, namely the electric field set up by the charge on the outer surface, goes off inside because of the distribution geometry. Now you can see that this field is not balancing anything but itself inside! The overall effect of the applied external field is to redistribute the charge on the outer surface in such a way that cancel the applied external field inside the conductor.

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  • $\begingroup$ Your answer is just a bit hard to understand $\endgroup$
    – lee
    Commented Jan 22, 2021 at 13:51
  • $\begingroup$ What is not clear? (note that I'm not an english native speaker) $\endgroup$
    – No Signals
    Commented Jan 22, 2021 at 13:58
  • $\begingroup$ Neither am I a native. What I didn't get is - "Are the first two to cancel each other on all the exterior points of the inner surface!" Did you mean that the first two fields will cancel each other. I have heard of the usage "Were you to..." but haven't heard "Are you to..." $\endgroup$
    – lee
    Commented Jan 22, 2021 at 15:25
  • $\begingroup$ Yes, you got it, thanks for the suggestion. $\endgroup$
    – No Signals
    Commented Jan 22, 2021 at 16:21
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This may be bs but if the field outside is independent of the distribution of charges inside then I'm guessing that 1) there exists a chg distib on the surface of the hole that makes the field outside the hole 0. Separately if one assumes that for the same conductor volume without the hole (ie if the condctr didn't have the hole) there is chg distrib that balances the field outside in the sense that it makes the field inside 0. Now this is just a crazy guess up to now, but if it works and makes all the boundary cond true thn our guess is the rigt sol n were done. Hmmm now is it?

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