## 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

its not a holow tube.

2. Disintegrati Group Title

3. sugargurl Group Title

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.

4. Disintegrati Group Title

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.

5. sugargurl Group Title

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..:)

6. ArchiePhysics Group Title

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.

7. Soham051994 Group Title

actualy E= du/ds. or can be expressed in terms of flux too

8. ArchiePhysics Group Title

Right, your definition is more general one.

9. jit4won Group Title

Use Gauss Law

10. Disintegrati Group Title

its not possible to use gauss law

11. jit4won Group Title

Is it restricted ?

12. Disintegrati Group Title

no, you cannot find the answer using gauss law, at least i think so..

13. ArchiePhysics Group Title

Sorry, I agree.

14. Disintegrati Group Title

sorry, it's E=(-dV)/ds

15. jit4won Group Title

E= $\sigma/\epsilon$

16. jit4won Group Title

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 <<<<<<

17. jit4won Group Title

l = side of the cube

18. Disintegrati Group Title

when you say that the field is independent of distance, do you mean the distance from the centre of the cube?

19. jit4won Group Title

yes

20. Disintegrati Group Title

in that case, |dw:1337706464916:dw|

21. Disintegrati Group Title

@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

22. egenriether Group Title

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.

23. or_dobkowski Group Title

where is the charge located? Or is it a genrall statement, just inside the cube

24. Disintegrati Group Title

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.

25. shivam_bhalla Group Title

@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.

26. shivam_bhalla Group Title

*cube

27. shivam_bhalla Group Title

*redistribute

28. Disintegrati Group Title

but i want to find the electric field on the surface of the cube.

29. sugargurl Group Title

thanks disintegre for correcting me...:)

30. yakeyglee Group Title

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.