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Disintegrati
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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
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

This Question is Open

binary3i Group TitleBest ResponseYou've already chosen the best response.0
its not a holow tube.
 2 years ago

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
i know, what's your point?
 2 years ago

sugargurl Group TitleBest ResponseYou've already chosen the best response.1
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

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
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

sugargurl Group TitleBest ResponseYou've already chosen the best response.1
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

ArchiePhysics Group TitleBest ResponseYou've already chosen the best response.0
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

Soham051994 Group TitleBest ResponseYou've already chosen the best response.0
actualy E= du/ds. or can be expressed in terms of flux too
 2 years ago

ArchiePhysics Group TitleBest ResponseYou've already chosen the best response.0
Right, your definition is more general one.
 2 years ago

jit4won Group TitleBest ResponseYou've already chosen the best response.0
Use Gauss Law
 2 years ago

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
its not possible to use gauss law
 2 years ago

jit4won Group TitleBest ResponseYou've already chosen the best response.0
Is it restricted ?
 2 years ago

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
no, you cannot find the answer using gauss law, at least i think so..
 2 years ago

ArchiePhysics Group TitleBest ResponseYou've already chosen the best response.0
Sorry, I agree.
 2 years ago

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
sorry, it's E=(dV)/ds
 2 years ago

jit4won Group TitleBest ResponseYou've already chosen the best response.0
E= \[\sigma/\epsilon\]
 2 years ago

jit4won Group TitleBest ResponseYou've already chosen the best response.0
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

jit4won Group TitleBest ResponseYou've already chosen the best response.0
l = side of the cube
 2 years ago

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
when you say that the field is independent of distance, do you mean the distance from the centre of the cube?
 2 years ago

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
in that case, dw:1337706464916:dw
 2 years ago

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
@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

egenriether Group TitleBest ResponseYou've already chosen the best response.0
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

or_dobkowski Group TitleBest ResponseYou've already chosen the best response.0
where is the charge located? Or is it a genrall statement, just inside the cube
 2 years ago

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
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

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@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

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
*cube
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
*redistribute
 2 years ago

Disintegrati Group TitleBest ResponseYou've already chosen the best response.0
but i want to find the electric field on the surface of the cube.
 2 years ago

sugargurl Group TitleBest ResponseYou've already chosen the best response.1
thanks disintegre for correcting me...:)
 2 years ago

yakeyglee Group TitleBest ResponseYou've already chosen the best response.0
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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