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choupi

Is voltaege zero for a conductor in electrostatic eq? I was thinking no, but since V=Ed it seems like there could be a difference in V by virtue of the particles being a certain distance apart. Could anyone explain Thanks!

  • one year ago
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  1. ghazi
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    @choupi it's change in voltage = E.change in distance that is \[dv=E.dx or E= dv/dx\] so what you are asking is bit unclear...but as far as i can understand it's just drifting of electrons under the influence of electric field that gives change in potential

    • one year ago
  2. r0oland
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    The electric field inside a conductor is always zero (for electrostatics), see: http://ocw.mit.edu/courses/physics/8-02sc-physics-ii-electricity-and-magnetism-fall-2010/conductors-and-insulators-conductors-as-shields/MIT8_02SC_notes9.pdf "Module 9: Conductors and Insulators" on page 16. Hence, the potential difference in that conductor is always zero. The whole conductor can however have a potential difference in respect to a reference point. This depends on the reference (eg. infinity where E=0) as well as the charge on and the geometry of the conductor.

    • one year ago
  3. Kumar777
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    good question E=-dv/dx since the electricfield inside the coductor is always zero you can make out that the potential inside a conductor is always constant.IT NEVER CHANGES.1 conductor 1 potential.Therefore potential difference inside a conductor is always zero.

    • one year ago
  4. harshaash
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    take a gaussian surface,just inside the conductor.now gauss law says \[\int\limits_{?}^{?}E.dA=(Const)q _{enclosed}\]and Q enclosed is 0 we have E is 0.now all the conductors charges lie on the outer surface,as they would lie max dist apart...is this the answer to your ques?well there could arise a question from this reasoning...ie..we proved that.\[\int\limits_{?}^{?}E.dA\] is 0 but it may not necessarily imply E is 0....but that can also be proved as E is a consv field.....

    • one year ago
  5. harshaash
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    it was not a contradiction..in fact your questing was not clear do u mean outside the conductor or inside? outside bearing no ext charges,E also 0 for the same reason..gauss law(when i meant,in the above reasoning that,take a gaussian surfacejust inside the conductor it means just inside its surface) but if there were ext charges E MAY not be 0

    • one year ago
  6. harshaash
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    well a practical application of above is..... u r advised to sit in cars on a rainy day,as any amount of thunder,may not effect you as you are in a closed metallic shell

    • one year ago
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