Electric field inside a uniformly charged cubical box

I have read that electric field at any point( not only at center) inside a uniformly charged spherical shell is zero(by symmetry).
But if we take a uniformly charged cubical box, will electric field be zero at every point inside the box?
I am confused plz help ( I think it should not be zero at every point)
thanks!

blue_leaf77
Homework Helper
I have read that electric field at any point( not only at center) inside a uniformly charged spherical shell is zero(by symmetry).
It's zero by Gauss law and the charge distribution on the shell need not be uniform.
EDIT: I think the shell should indeed be uniform.
But if we take a uniformly charged cubical box, will electric field be zero at every point inside the box?
What kind of box is it, a solid one or the hollow one?

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• Majorana
blue_leaf77
Homework Helper
it's a hollow cubical box
Then you can use Gauss law.

• Majorana
blue_leaf77
Homework Helper
electric flux =0 does not necessarily imply E=0
You are right, but in this particular problem, the way you solve for the field is by looking at the fact that no matter which choice of the enclosing surface (of which there are infinite number of them) you apply for the flux integral (so long as it's contained inside the box), the total flux is still zero. From this you can conclude that the field is zero.

EDIT: On a second thought, I got twisted with the fact that the system does not exhibit spherical symmetry (which I have assumed unwittingly). I am not sure if my argument above can be used to explain this thing.
I would recommend to wait for other members to share their thought.

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I have read that electric field at any point( not only at center) inside a uniformly charged spherical shell is zero(by symmetry).
But if we take a uniformly charged cubical box, will electric field be zero at every point inside the box?
I am confused plz help ( I think it should not be zero at every point)
thanks!
It's true that the net flux for a closed surface inside a hollow cube would be zero (since q enclosed=0), but that doesn't imply that the electric field is also zero at every point inside the cube. The very thing that makes Gauss' law useful is when by symmetry the 'E' comes out of the surface integral, you cannot do this for the cube since we have no reason to believe that the E field is constant at the entire Gaussian surface..
For a sphere (this property that the E field is zero inside a charged shell, is called the shell theorem) the case is different.. Here each point on the Gaussian surface is identical to the others, so we can say that the E field must be constant throughout the surface ( and the 'E' part comes out of the integral).

• ajeet mishra
Also there's a geometrical property of spheres by which you can prove the shell theorem without using Gauss' law ( cubes don't have this property).. See Pg4, Chapter 5 of Feynman Lectures, Vol.2

• ajeet mishra
Henryk
Gold Member
First, what do you mean by
a uniformly charged cubical box,
If you mean that the there is a uniform charge density on all the faces of the box, then the answer to your question is, no, the electric field will not be zero at every point.
If, however, you mean the faces of the box are charged to the same potential (e.g. box made of metal), then the answer is yes, the electric field will be zero everywhere.
For any enclosed volume with a constant potential at the surface, the potential inside has to be constant and equal to that at the surface. You can prove it easily by assuming the potential inside the box can vary. Then there has to be a point of either maximum or minimum potential. At that point, ## \nabla ^2 V ## is not zero and has to be equal to ## \rho / \epsilon _0 ##. But it is not possible since there is no charge inside. Hence the potential has to be constant. The electric field, ## E = - \nabla V ## is, therefore zero.

• ajeet mishra and beamie564
Then there has to be a point of either maximum or minimum potential. At that point, ## \nabla ^2 V ## is not zero and has to be equal to ## \rho / \epsilon _0 ##. But it is not possible since there is no charge inside. Hence the potential has to be constant. The electric field, ## E = - \nabla V ## is, therefore zero.[/QUOTE]
Thank you sir!
And what I have concluded is that
Electric field inside any three dimensional hollow geometrical shape will be zero irrespective of their charge distribution(by the same argument).
Is it correct?

No.. This is true only for conducting materials ( metals).. It's not a general result

Henryk
Gold Member
Electric field inside any three dimensional hollow geometrical shape will be zero irrespective of their charge distribution(by the same argument).
Is it correct?
No, this is not true. What is true is that the electric field will be zero if and only if the potential at all the faces is constant. To get a constant potential, there is only one particular surface charge distribution that will give you that. The actual distribution depends strongly on the shape. For a sphere, the charge distribution is uniform but for other shapes it is not.

• ajeet mishra
. At that point, ∇2V∇2V \nabla ^2 V is not zero and has to be equal to ρ/ϵ0ρ/ϵ0 \rho / \epsilon _0 . But it is not possible since there is no charge inside. Hence the potential has to be constant.
This came to my mind suddenly, Laplacian(V)= 0 doesn't necessarily mean that V is a constant..Eg. for V=2xy
Am I wrong?

• ajeet mishra
Henryk
Gold Member
This came to my mind suddenly, Laplacian(V)= 0 doesn't necessarily mean that V is a constant..Eg. for V=2xy
Am I wrong
You are absolutely right. Take any example of a field from a physics textbook, like a field of a point charge. Outside the charge, the Laplacian is zero, yet the potential goes like ## \frac1 r ##. Or a field between the plates of a capacitor, etc, etc.
But where the potential has a local minimum or a maximum, then the Laplacian is non-zero. Let's take a minimum for example. The potential around minimum can be expanded as ## V(x,y,z) = V_{min} + a* x^2 + b * y^2 + c * z^2 ## where a, b, c are all positive. The Laplacian is simply ##2 *(a +b +c) \neq 0 ##
The whole point of my argument was that inside the box with walls at the constant potential, there cannot be a local minimum or maximum, therefore, the potential has to be constant and exactly the same as at the walls.

• Rodrigo Simoes and ajeet mishra