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The reasoning provided behind why the electric field inside a metallic conductive sphere is zero, in my textbook is - "In case of a metallic (conducting) sphere, the entire charge will reside on the outer surface of the sphere. Therefore there will be no charge within the gaussian surface. In that case by Gauss's theorem we get The net flux inside = 0"

I kind of understand and at the same time do not understand this. I can't quite phrase the question so am unable to get the answer by googling. The charge outside the surface can't flow to the inside surface? Conductive material carry charge, so why is that in case of a sphere the charge is just staying on the outer surface isolated from inside?

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    $\begingroup$ Charge moves from higher potential to lower potential. Since when charges are outside , they are far from each other and so less potential. So it will in any case move from inside(higher potential) to outside (lower potential) $\endgroup$ Sep 5, 2022 at 11:02
  • $\begingroup$ [Trivial conclusion once one learns this ](physics.stackexchange.com/questions/611370/…) $\endgroup$ Sep 6, 2022 at 15:42
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    $\begingroup$ Is this a hollow sphere, or solid? (Technically, a sphere is hollow; if it's solid, it's a ball.) $\endgroup$
    – gidds
    Sep 7, 2022 at 17:37
  • $\begingroup$ When I first read the question I was thinking, "What if a metallic sphere is constructed around some pre-existing charge?" $\endgroup$
    – Michael
    Sep 7, 2022 at 18:53

6 Answers 6

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The no net volume charge within a conductor has gained the legend status level, but things are, as usual, more complex.

Combining the charge conservation, Ohm's law and Maxwell's second equation, one gets:

$$\begin{cases} \frac{\partial \rho }{\partial t} + \overrightarrow{ \nabla }. \overrightarrow{j} =0 \\\overrightarrow{j}= \sigma \overrightarrow{E} \\\overrightarrow{ \nabla }.\overrightarrow{E} = \frac{ \rho }{ \varepsilon _{0}} \end{cases} ~~\Rightarrow ~~ \frac{\partial \rho }{\partial t}+\frac{ \sigma \rho }{ \varepsilon _{0}}=0~~ \Rightarrow ~~\rho(t)=\rho(0)e^{-\frac{ \sigma }{ \varepsilon _{0}}t }$$

Wikipedia gives for copper:$$\sigma=5.96×10^{7}~~S.m^{-1}~~at~~20~~°C.$$ $$\varepsilon _{0}= 8.85×10^{-12}~F⋅m^{-1}$$

So: $\frac{ \sigma }{ \varepsilon _{0}} \approx 6.7~10^{18}~ s^{-1}$

The time $\triangle t$ for 99% of $ \rho _{0}$ to diffuse to the surface is: $$ \triangle t =- \frac{\ln(0.01)}{6.7~10^{18}} \approx 7.10^{-19} s$$

So one can safely says there are no charges within a perfect conductor.

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    $\begingroup$ It really looks too neat not to be presented in all the courses about Classical Electromagnetism! I'll think about it. Could it be so easy to explain? +1. I'll come back if I find any flaw in this process. Hope none, so that I have learnt something easy and new $\endgroup$
    – basics
    Sep 5, 2022 at 19:43
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    $\begingroup$ @Basics At University, my professor Roger Petit, was the French equivalent of David Jackson in the US. He is the one to be credited to have taught me these lines. $\endgroup$
    – Shaktyai
    Sep 5, 2022 at 21:41
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    $\begingroup$ @Shaktyai is some of his material like lecture notes and such avaliable anywhere? $\endgroup$ Sep 6, 2022 at 19:04
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    $\begingroup$ @basics Ondes électromagnétiques en radioélectricité et en optique -Roger Petit-1993 (In French). that is basically the notes from his courses. Electromagnetic Theory of Gratings -R. Petit (auth.), Professor Roger Petit (eds.)-1980 . Give me your email if you do not find them. I will emailed them to you. $\endgroup$
    – Shaktyai
    Sep 6, 2022 at 19:42
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    $\begingroup$ Can you include the units? E.g., near "1900". How can the unit of the end result be seconds? Where did "F" go? $\endgroup$ Sep 6, 2022 at 21:45
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There can be an electric field inside a conductor, but that exists only for a short time (unless you are using a battery to maintain the field).

Now conductors (let's assume a spherical one) have free electrons which can move if an electric field is applied. So when you give some charge (let's assume to be positive) to the conductor, it creates an electric field and so electrons will move towards that region and this movement continues until the electrons came from the outer surface. Once the outer surface is charged there could be no more flow as there is no more electric field inside the conductor since now the charged sphere is similar to a uniformly charged shell.

So ultimately all the charges goes to the outer surface of the conductor (of any shape).

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If there were charges inside then they'd repel each other, until the charge goes to the surface. Here the force from repulsion equals the force from the molecules at the surface (molecules of the conductor with which the charge is connected to). This is also a general result, irrespective of the shape of the conductor.

What about the case when we have an external field too? We know that a surface charge will be induced, causing the field to go zero in the interior of the conductor. If we now add an extra charge then the conditions are identical to the case when the external field is zero. In other words, the charge again flows to the surface.

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A conductor filling the volume $V$, by definition, contains free charge carriers, let's say of charge $q$. Now suppose that there is a region $U \subset V$ within the conductor in which the charge density is $\rho \neq 0$, then Gauss's law states that there will be an electric field component perpendicular to the surface $\partial U$ of $U$, which exerts a force on the free charge carriers and causes them to move out of or into $U$ (depending on the sign of $q \rho$) until $\rho = 0$ at every point $p$ inside $V \setminus \partial V$. There follows, that in electrostatics the inside of any conductor is free of electric fields and non-zero charge densities.

On the other hand, for $p \in \partial V$ it may be $\rho(p) \neq 0$, because the electric field can be perpendicular to $\partial V$, in which case the charge carriers cannot move to reduce it, since they cannot leave $V$. This implies, that surface charges can exist, if without them there would be a non-zero electric field within $V$, caused for example by charges outside of $V$ or by non-zero total charge of $V$.

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The question a static problem that is ambiguous as it is stated because it is not clear with regard to charge that may be present on the inside surface of the conductive sphere. There can be a charge in the interior of the inner surface of the sphere, and this charge is neutralized by an opposite charge on the interior surface of the conductive sphere distributed so that there is no electric field in the interior of the conductor. There is a charge equal to the charge on the exterior surface of the sphere equal to the charge in the interior. So the charge enclosed by the outer surface of the sphere, not counting the charge on the outer surface, is zero, but the charge enclosed by the inner surface of the sphere, not counting the charge on the inner surface of the sphere, need not be zero.

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The electric field inside a conductor is always 0.

Let's take a conductor which has neutral charge, i.e., an equal amount of positive and negative charge, and has a test positive charge outside a shell.

Now due to this test charge it has some electric field. Now, when this electric field passes through the conductor, the charges present in the conductor aligns them, in such a way that the negative charge will be towards the positive test charge (because of attraction) and the positive charge repelling the test charge. You may see that the charges will reside on its surface.

Actually an internal electric field is created in an conductor.

Reference

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    $\begingroup$ E always 0 in conductor negates the existence of Ohm's law. In a material (including metals), $\vec J =-\sigma \vec E$ where $\vec J$ is the current density, sigma is the electrical conductivity and E is the electric field. $\endgroup$ Sep 6, 2022 at 18:53
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    $\begingroup$ Are we to guess that "hv" means "have" (or rather "has"), "+ive" means "positive", and "-ive" means "negative"??? What is up with that? Where is that commonplace? $\endgroup$ Sep 6, 2022 at 21:58
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    $\begingroup$ The last sentence is incomprehensible. $\endgroup$ Sep 6, 2022 at 21:59
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    $\begingroup$ The question was why the CHARGE inside a solid conducting sphere is zero. You haven't addressed that question; although you've made two other claims, which may or may not be related to the question. You first claimed that the electric FIELD inside a sphere is zero; then you claimed (if I understand your last sentence correctly) that there can be an electric FIELD inside a sphere. In other words, your last claim contradicts your first one; and neither claim answers the question. Also, English is my first language and I'm having difficulty making sense of parts of this. $\endgroup$ Sep 8, 2022 at 5:54
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    $\begingroup$ Yes my friend the last sentence actually is related with the upper part because when the charges align themselves in such way (refer to image) then it is obvious that the internal electric field is there too $\endgroup$ Sep 8, 2022 at 6:28

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