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Field inside Conductor |
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| Aug11-05, 01:44 AM | #1 |
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Field inside Conductor
Field inside conductor in electrostatic is zero. So they say...cuz the free-particle will move due to eny Electricfield present until they all stablized to the surface of the conductor to cancel the Electricfield present inside, thereby making the net field inside zero.
After understading the argument such as above, I set my work to show the electric field inside conductor is zero(Given total charge inside of the conductor but not the surface is zero) rather mathematically and got nowhere. Any help or advise will be appreciated. If I already know total charge inside of the conductor is zero, from Gauss's law it follows that Electric field inside of conductor must be constant value. So it means that it must be either zero or non-zero constant value. Now, how can I show that zero is the only value possible? |
| Aug11-05, 04:01 AM | #2 |
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If the electric field inside the conductor were a nonzero constant then electrical current (from both positive and negative charges) would flow to the surface and charge would build up indefinitely.
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| Aug11-05, 11:27 AM | #3 |
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If there were any field the free charges would move and there wouldn't be Electrostatics anymore. The induced charges play the role here,the field due to them tends to cancel off the original field. When you put conductor in an external electric field,this drives the positive and negative charges apart like positive to right and negative to left if E is to the right.Hence an electric field develops which is equal to the external electric field.That Leads to ZERO. There is no such great mathematics involved here,as you can see by formula of induced charges it is exactly equal and opposite to charge producing external electric field.This looks simple i guess. [tex] \oint{\vec{E} \cdot d\vec{a}} = \frac{Q_{enc}}{\epsilon_0} [/tex] There is no constant of integration here! |
| Aug11-05, 01:48 PM | #4 |
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Field inside Conductor[tex]\oint{\vec{E}\cdot d\vec{a}} = 0 [/tex] doesn't say that [tex] \vec{E} = 0 [/tex]. It could be that E is zero or E dot dA is zero. For example, if the Electric field is constant over the region of Gaussian surface then we still have [tex]\oint\vec{E}\cdot d\vec{a}} = 0}[/tex]. I was wondering if one can show explicitly(from math point of view) that [tex] \vec{E} = 0[/tex] is necessary result. I didn't say the total charge is zero, I said the charge inside of the gaussian surface which in this case the entire volume of the conductor except at the surface of conductor itself the charge is zero. |
| Aug11-05, 02:23 PM | #5 |
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| Aug11-05, 07:33 PM | #6 |
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Can you explain further? If the electric field is constant in magnitude and direction, the above closed surface integral will still give zero. That is why I mentioned that E could be either zero or constant value. What makes you say that E could be zero only? |
| Aug11-05, 09:02 PM | #7 |
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If there is an E field present, currents will flow within the conductor until the E field is zero.
There is no mathematical argument as to why E must be zero, only physical ones. For example, in the case of an insulator, the situation is different, despite the fact the same equation is involved. Claude. |
| Aug11-05, 11:21 PM | #8 |
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| Aug12-05, 07:48 AM | #9 |
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) Claude Bile is correct; the argument for E = 0 within a conductor is a physical one. |
| Aug12-05, 10:22 AM | #10 |
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After all, I dont speak English well. Thanks for your reply! I knew the physical explanation since that was only explanation I could find from Grifitth and Feynmman but I thought that they were skipping the mathematical arguement. But still, it doesn't quiet make sense to me. So if I put a single particle with some charge inside conductor, that particle will not accerlate because there are no electric fields present inside of conductor? But for the other particles, now there are electric field inside conductor due to the particle that I just put in? |
| Aug12-05, 10:29 AM | #11 |
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Divergent of E is equal the total Charge inside. (Gauss's law) If the total Charge inside is zero(since that is what I am assuming) the Divergent of E must also be equal to zero. Which tells me that E must be constant (whether zero or not) I am sorry for not being able to write clearly, I was simply trying to derive the result that inside of conductor Electric field must be zero given that charge inside is zero. Most of the texts that I've been reading along with do this conversly. They say E field inside of conductor is zero then derive the result that the charge inside of conductor must also be zero. Like above people point out, maybe there can't be given a mathematical arguement to show Electric field must be zero in that closed surface integral being equal to zero. Only the physical one. But there might be just one, one can not know such thing for sure until he shows that it is impossible to show. Right?
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| Aug12-05, 11:56 AM | #12 |
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| Aug12-05, 02:09 PM | #13 |
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Say I have a conductor of spherical shell. We claim that inside of such conductor the electric field is zero. If I can put an electron inside of such conductor, surely the electron will not distribute itself because there isn't any Electric field to move such electron since it is a conductor. So, I imagined that the electron will just sits inside of conductor where I put it. Then this conductor is not a conductor anymore becouse now there's electric field insdie due to the electron I just put it in. Is this a total nonsese? |
| Aug12-05, 02:24 PM | #14 |
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If you could magically fix an extra electron inside a conductor, then the other charge would just redistribute themselves to cancel the field. |
| Aug12-05, 03:33 PM | #15 |
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If you put an electron into a conductor, its electric field would polarize the charge distribution on the surface of the conductor. This redistributed charge would attract the electron to the surface of the conductor in about 10^-19 seconds, leaving no field and no electron inbside the conductor.
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| Aug12-05, 09:58 PM | #16 |
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V = d/t = 0.1m/10^-19 sec. V = 10^18 meters/sec. Hmm; not bad for those of us trying to create superluminal electrons.
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| Aug13-05, 06:59 AM | #17 |
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Don't think of that single electron as working its way to the surface at superluminal speed. Instead, think of the conductor being filled with a "sea of electrons" which shift ever so slightly, which happens very quickly. The net result is that the surface of the conductor has gained one electron's worth of charge (but the electron placed in the middle doesn't have to travel to the surface).
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