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We have studied so far that electric field inside a conductor if no charge is placed inside is zero. But we know that every conductor has only a limited number of electrons. What happens when ALL the electrons have aligned to cancel the field inside conductor under application of external field and then we increase the field a little bit more certainly there are no more electrons to cancel the field and this additional field must be present inside the conductor now.

I know that practically under electric fields of magnitude that can drag out all electrons from the conductor's body to surface field emission and/or electrical breakdown would be taking place, But is this phenomenon theoretically possible ?

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The maximum electric field strength that the electron sea can set up in a metal to oppose an external field is the same field strength that results in ionisation--in which case the electrons have left the metal. If you want to do the calculations, consider the maximum field that a pair of electrons can set up on a small distance about the diameter of a typical metal atom. It is way higher than the ionisation field strength.

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  • $\begingroup$ I did calculations for a shell and there is direct dependence on r for electric field. Acc. to those calculations we could test the hypothesis at nano scale but how tl measure field inside nano sized metalic shells is a problem. $\endgroup$ – Rijul Gupta Oct 20 '13 at 21:22
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We would require an extremely high intensity electric field to make the conductor 'run out' of electrons. As we know, the external electric field is countered by electrons arranging themselves, thereby creating another electric field (keep this in mind, since it means that at the very moment the field is applied, it does in fact penetrate the conductor, but this changes very fast).

When the field magnitude is very high, the effects you spoke of can theoretically happen. At the very least though, the intrinsic electric field established by the movement of electrons will stay constant.

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  • $\begingroup$ actually via some calculations I was able to see that we either need a spherical conducting shell of radius in nano metres or a field of order 10^10. the field is definitely going to cause electrical breakdown, but I found that in nano scale it was observable with smaller fields which would not cause breakdown. The problem is how would we measure the field inside a conductor or conducting shell at micro scales ? $\endgroup$ – Rijul Gupta Oct 20 '13 at 21:00
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diagram

Note : Extremely Approximated calculations just to check at which scale the effect may be observable

Approximation : Somhow got external field along a single line and all the charges of conductor came to surface equally to two halve rings in the plane of electric field "line" to cancel the external field.

We know that field of a semicircular ring due to charge Q is directed radially away from ring from the centre and has magnitude given by : 
Ein = 2kQ/(PI×R^2)

So net electic field due to charges accumulated along ring would be

Ein = 4kQ/(PIxR^2)

Qmax = (total no of electron)×(charge of electron)
Qmax = (moles of metal) × (atomic number) × e
Qmax = (Volume of conductor) × (density) × (avogadro no.) × Z × e / (atomic mass)
Qmax = (4/3 × PI×R^3) × RD × Na × Z × e / M

Emax = (4×k/PI×R^2) × Qmax
Emax = (4×k/PI×R^2) × (4/3 × PI × R^3) × RD × Na × Z × e / M
Emax = (16/3 × 9 × 10^9 × R) × RD × 6.022 × 10^23 × Z × 1.6 × 10^-19 × / M
Emax = (4.6×10^15  × RD × Z/M) × R

for radius in nano scale we will get electric field that would not cause electrical breakdown, on this scale or maybe smaller the effect can be possibly observed. Otherwise the large field would just continue to cause electric breakdowns.

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We have studied so far that electric field inside a conductor if no charge is placed inside is zero.

False: the electric field inside a conductor is zero as long as the situation is static. No variable external fields -> $\vec{E}(inside) = 0$. Otherwise, with $\vec{E}(inside) \neq 0$ you would have a electrons moving inside the conductor. Then your static hypothesis would be false. Note that this includes the case where you place a charge inside the conductor. It would just move and rearrange the other electrons in order to re-set $\vec{E}(inside) = 0$.

This can happen as long as $\int_{S} \vec{\sigma} \cdot \vec{dA} < Q_{free}$ where $S$ is the conductor's surface and $Q_{free}$ is the total conductor's free charge.

(From the electric boundary condition you get $\vec{E}_{out} - \vec{E}_{in} = \vec{E}_{out} - 0 = \frac{\sigma}{\epsilon_0}$. Integrating $\sigma$ over the conductor's surface you get the total polarization charge.)

But we know that every conductor has only a limited number of electrons. What happens when ALL the electrons have aligned to cancel the field inside conductor under application of external field and then we increase the field a little bit more certainly there are no more electrons to cancel the field and this additional field must be present inside the conductor now.

This happens when $\int_{S} \vec{\sigma} \cdot \vec{dA} = Q_{free}$.

What happens now? Simple: from the electric boundary condition you still get that $\vec{E}_{out} - \vec{E}_{in} = \frac{\sigma}{\epsilon_0}$, but this time $\vec{E}_{in} \neq 0$. That is, you will get a static electric field inside that is not zero. Note that now the conductor is no longer a conductor, since you imposed that all the free charge resides on the surface (only the surface is still a conductor, not the inside).

I know that practically under electric fields of magnitude that can drag out all electrons from the conductor's body to surface field emission and/or electrical breakdown would be taking place, But is this phenomenon theoretically possible?

There is $\sigma_{max}$ achievable, depending on the conductor's composition. It is explained by the fact that electrons bounding energy in metals is finite. When the charge density is high enough, external electrons will be pushed outside of the conductor since the force they experiment on themselves is higher the boundary force plus the external force field.

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consider a sphere which is kept in electric field.. so negative charge is induced on one side and +ve on another side. if all the free electrons are ionised and still there is electric field inside a conductor then it means "E" inside a conductor is pointing from negative to positive which is a violation of direction of E field.so my guess is atoms will be polarised to make 0 E field or will break the bonds like high current passing through wire

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A typical small wire contains thousands of coulombs of electrons in the conduction band. For example, in Cu metal, each atom donates one mobile electron to the sea-of-charge.

Copper has 8.5e22 atoms/cc, and with electron charge 1.6e-16 coulombs, copper's "electron sea" has 13,600 coulombs/cc.

Even a simple circuit might have one or two cubic centimeters of copper. Assume that the circuit as a whole, treated as an isolated sphere, has a capacitance of a couple of picofarads.

V = Q/C = 1.4e4 coul / 1e-12 F = 1.4e16 volts

energy = .5 C V^2 = .5 * 1e-12 * 2e32 = 1e20 joules

energy contained in one ton of TNT = 4.2e9 joules

energy required to remove electrons from small circuit = 1e20 / 4.2e9 = 24,000 MEGATONS OF TNT

So, don't remove all the electrons from a resistor. They'd spark! After things settled down, the end result would be a vast pool of lava centered on your neighborhood, with several burning cities around the (distant) crater edges.

KLINGONS TRY TO REMOVE ALL THE ELECTRONS FROM A SMALL ROCK?

On the other hand, doped semiconductor typically has one mobile charge-carrier per 10^6 to 10^9 atoms, and in diodes, BJTs and FETs, all the charges are swept out of the microns-thick layer called "depletion zone." That's the central 'magic' behind semiconductors: a charge density so low that it becomes easy to sweep all the electrons out of a thin layer, converting that layer into insulator.

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