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Disintegrati Group Title

suppose i have a conducting cube of side length 1 metre, and a charge q is placed inside it. how can i find the electric field at a point which is at the centre of any face of the cube? (the charge eventually gets dispersed to the surface of the cube, as it is a conductor)

  • 2 years ago
  • 2 years ago

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  1. binary3i Group Title
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    its not a holow tube.

    • 2 years ago
  2. Disintegrati Group Title
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    i know, what's your point?

    • 2 years ago
  3. sugargurl Group Title
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    electric field=qd where d is the distance and q is the charge..:) distance can be taken as lenght of the conducter and charge q is the charge of electron or proton.

    • 2 years ago
  4. Disintegrati Group Title
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    yes, but the charge will get distributed unevenly on the surface of the cube, that's why i'm having a problem finding the field.

    • 2 years ago
  5. sugargurl Group Title
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    when charge is gets unevenly distributed each charge creats its own electrical field that field is actually equivalent to electric field of a conducter... as electric field is a scalar quantity..:)

    • 2 years ago
  6. ArchiePhysics Group Title
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    Electric field is a vector quantity, and it is not equal to qd it is E = U/d where U is potential difference and d is the distance between two points having these potentials.

    • 2 years ago
  7. Soham051994 Group Title
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    actualy E= du/ds. or can be expressed in terms of flux too

    • 2 years ago
  8. ArchiePhysics Group Title
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    Right, your definition is more general one.

    • 2 years ago
  9. jit4won Group Title
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    Use Gauss Law

    • 2 years ago
  10. Disintegrati Group Title
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    its not possible to use gauss law

    • 2 years ago
  11. jit4won Group Title
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    Is it restricted ?

    • 2 years ago
  12. Disintegrati Group Title
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    no, you cannot find the answer using gauss law, at least i think so..

    • 2 years ago
  13. ArchiePhysics Group Title
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    Sorry, I agree.

    • 2 years ago
  14. Disintegrati Group Title
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    sorry, it's E=(-dV)/ds

    • 2 years ago
  15. jit4won Group Title
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    E= \[\sigma/\epsilon\]

    • 2 years ago
  16. jit4won Group Title
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    In this case Each surface will act like charged conducting surface . The electric field near a charged surface is independent of distance . You can use the same formula to calculate E at the center . The formula is good if l/r <<<<<<

    • 2 years ago
  17. jit4won Group Title
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    l = side of the cube

    • 2 years ago
  18. Disintegrati Group Title
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    when you say that the field is independent of distance, do you mean the distance from the centre of the cube?

    • 2 years ago
  19. jit4won Group Title
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    yes

    • 2 years ago
  20. Disintegrati Group Title
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    in that case, |dw:1337706464916:dw|

    • 2 years ago
  21. Disintegrati Group Title
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    @jit4won , i agree to what you say, but that only applies to gaussian surfaces which entirely cover the conductor. so, if i use the gauss law in this case, i'll have to take the radius of the sphere equal to the distance from the centre of the cube to one of its corners. there's no point, as the point that i need the electric field at is not included on the periphery of the gaussian surface

    • 2 years ago
  22. egenriether Group Title
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    Also remember the field just outside the box will be perpendicular to the surface of the box (i.e. parallel to the area vector) This is a requirement, all electric field lines going into a conductive surface curve into the surface to be perpendicular.

    • 2 years ago
  23. or_dobkowski Group Title
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    where is the charge located? Or is it a genrall statement, just inside the cube

    • 2 years ago
  24. Disintegrati Group Title
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    when you put a charge inside a conducting surface (it may be placed anywhere inside it), it redistributes itself on the surface due to the repulsion. in case of a sphere, the charge distribution on the surface is uniform, but it's not in case of a cube.

    • 2 years ago
  25. shivam_bhalla Group Title
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    @Disintegrati, when charge q in placed inside the conducting xube, the charge redesrtibutes itself to the outer surface of the cube in such a way that the net electric field inside conductor is zero. So your answer should be simply zero.

    • 2 years ago
  26. shivam_bhalla Group Title
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    *cube

    • 2 years ago
  27. shivam_bhalla Group Title
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    *redistribute

    • 2 years ago
  28. Disintegrati Group Title
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    but i want to find the electric field on the surface of the cube.

    • 2 years ago
  29. sugargurl Group Title
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    thanks disintegre for correcting me...:)

    • 2 years ago
  30. yakeyglee Group Title
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    The electric field at the requested point is NOT zero, since we're looking at the surface. All i can tell you immediately is that he charge is going to collect at the corners...I still need to work this out for myself though.

    • 2 years ago
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