anonymous
  • anonymous
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)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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binary3i
  • binary3i
its not a holow tube.
anonymous
  • anonymous
i know, what's your point?
anonymous
  • anonymous
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.

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anonymous
  • anonymous
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.
anonymous
  • anonymous
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..:)
anonymous
  • anonymous
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.
anonymous
  • anonymous
actualy E= du/ds. or can be expressed in terms of flux too
anonymous
  • anonymous
Right, your definition is more general one.
jit4won
  • jit4won
Use Gauss Law
anonymous
  • anonymous
its not possible to use gauss law
jit4won
  • jit4won
Is it restricted ?
anonymous
  • anonymous
no, you cannot find the answer using gauss law, at least i think so..
anonymous
  • anonymous
Sorry, I agree.
anonymous
  • anonymous
sorry, it's E=(-dV)/ds
jit4won
  • jit4won
E= \[\sigma/\epsilon\]
jit4won
  • jit4won
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 <<<<<<
jit4won
  • jit4won
l = side of the cube
anonymous
  • anonymous
when you say that the field is independent of distance, do you mean the distance from the centre of the cube?
jit4won
  • jit4won
yes
anonymous
  • anonymous
in that case, |dw:1337706464916:dw|
anonymous
  • anonymous
@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
egenriether
  • egenriether
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.
anonymous
  • anonymous
where is the charge located? Or is it a genrall statement, just inside the cube
anonymous
  • anonymous
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.
anonymous
  • anonymous
@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.
anonymous
  • anonymous
*cube
anonymous
  • anonymous
*redistribute
anonymous
  • anonymous
but i want to find the electric field on the surface of the cube.
anonymous
  • anonymous
thanks disintegre for correcting me...:)
anonymous
  • anonymous
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.

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