Electric field and potential at vertex of cube

In summary, the smaller cube has a field and potential that is half of the field and potential of the larger cube.
  • #1
Krushnaraj Pandya
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Homework Statement


A solid non-conducting cube of side l and uniformly distributed charge q, has electric field E and potential V at one of its vertex, imagine this cube to be made of 8 smaller cubes of side l/2. if one such cube is removed, find the new field and potential at the point where the vertex used to be.
2. The attempt at a solution
Intuitively i thought, if E is field due to a cube of side l and charge q, the smaller cube has side l/2 and charge q/8. usually electric field is proportional to q and inversely proportional to r, assuming the same we get the field due to smaller cube E/2 which seems fair since it was much closer than the other cubes. by the same logic, potential usually falls of as 1/r, so potential due to smaller cube is V/4 but this looks a bit fishy...can someone show me the right way to look at this? I'll be really grateful
 
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  • #2
I think you are looking at it in the right way. I agree with your answers for the field and potential at a vertex of the smaller cube. Can you explain why it "looks a bit fishy" to you?

You didn't give your answer to the original question. But maybe that's because you see how to to get the answer from your results for the smaller cube.
 
  • #3
TSny said:
I think you are looking at it in the right way. I agree with your answers for the field and potential at a vertex of the smaller cube. Can you explain why it "looks a bit fishy" to you?

You didn't give your answer to the original question. But maybe that's because you see how to to get the answer from your results for the smaller cube.
my answers assumed that electric field is proportional to charge and inversely to length squared (I do not know if that is true for a cube, just intuition).
I realize now that there's nothing fishy about the answer, I was just confused whether a single cube (even when it is closer) can contribute a quarter of the potential. I realize now that it is quite logical. And yes I didn't give my answers to the original question since I only have to substitute the value found and subtract from the field and potential of the larger cube
This was largely based on intuition, can you tell me a more rigorous way to see it?
 
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  • #4
OK
Krushnaraj Pandya said:
my answers assumed that electric field is proportional to charge and inversely to length (I do not know if that is true for a cube, just intuition).
Did you mean to say that the field is inversely proportional to length squared?

The field and potential at a point can be considered as due to the superposition of the fields and potentials of all of the infinitesimal charge elements ##dq = \rho dxdydz## in the cube. Going from the large cube to the small cube is just a matter of rescaling distances.
(##\rho## remains the same.)
 
  • #5
TSny said:
OKDid you mean to say that the field is inversely proportional to length squared?

The field and potential at a point can be considered as due to the superposition of the fields and potentials of all of the infinitesimal charge elements ##dq = \rho dxdydz## in the cube. Going from the large cube to the small cube is just a matter of rescaling distances.
(##\rho## remains the same.)
yes, I meant squared...sorry I mistyped.
I understand that, that's how I derived my answer. what I meant was, is there any logical proof which can show that field is indeed proportional to Q and inversely to distance squared for a cube? I know its always true at large distances but this is a vertex!
 
  • #6
Krushnaraj Pandya said:
what I meant was, is there any logical proof which can show that field is indeed proportional to Q and inversely to distance squared? I know its always true at large distances but this is a vertex!
The field at some point p is the superposition of the fields due to all of the charge elements ##dq## in the cube. For each charge element ##dq##, the field at p is proportional to ##dq## and inversely proportional to the square of the distance between ##dq## and the point p. From this, you can construct a scaling argument for why the net field at a corner of the smaller cube is half the field at a corner of the larger cube.
 
  • #7
TSny said:
The field at some point p is the superposition of the fields due to all of the charge elements ##dq## in the cube. For each charge element ##dq##, the field at p is proportional to ##dq## and inversely proportional to the square of the distance between ##dq## and the point p. From this, you can construct a scaling argument for why the net field at a corner of the smaller cube is half the field at a corner of the larger cube.
Integrating these expressions often gives results different from that of a point charge, for example in the case of a ring...which is why I'm unsure if finally the field at p is proportional to ##q## and inversely proportional to the square of the distance.
 
  • #8
Krushnaraj Pandya said:
Integrating these expressions often gives results different from that of a point charge, for example in the case of a ring...which is why I'm unsure if finally the field at p is proportional to ##q## and inversely proportional to the square of the distance.
can you show me a rigorous proof for the following statement- "The sum of the electric field vectors due to the point charges add up to a function showing that total electric field is proportional to Q and inversely proportional to side length squared in case of a cube" ?
 
  • #9
Krushnaraj Pandya said:
can you show me a rigorous proof for the following statement- "The sum of the electric field vectors due to the point charges add up to a function showing that total electric field is proportional to Q and inversely proportional to side length squared in case of a cube" ?
If you are familiar with integrating over charge distributions to get the field, then you can examine the form of the integral for the field at the vertex of the cube in order to deduce that the resultant field at a vertex of the cube is proportional to Q and inversely proportional to L2. It is not necessary to actually carry out the integration.
 
  • #10
TSny said:
If you are familiar with integrating over charge distributions to get the field, then you can examine the form of the integral for the field at the vertex of the cube in order to deduce that the resultant field at a vertex of the cube is proportional to Q and inversely proportional to L2. It is not necessary to actually carry out the integration.
alright, thank you very much :D
 

1. What is an electric field at the vertex of a cube?

The electric field at the vertex of a cube is the force experienced by a unit positive charge placed at that particular point. It is a vector quantity and is represented by the symbol E. The direction of the electric field at the vertex is along the lines of the electric field that are tangent to the surface of the cube.

2. How is the electric field at the vertex of a cube calculated?

The electric field at the vertex of a cube can be calculated using the formula E = Q/4πεr², where Q is the charge of the cube, ε is the permittivity of the medium surrounding the cube, and r is the distance from the vertex to the center of the cube.

3. What is the potential at the vertex of a cube?

The potential at the vertex of a cube is the amount of work required to bring a unit positive charge from infinity to that particular point. It is represented by the symbol V and is a scalar quantity. The potential at the vertex of a cube is directly proportional to the electric field at that point.

4. How is the potential at the vertex of a cube related to the electric field?

The potential at the vertex of a cube is directly related to the electric field at that point. The potential can be calculated using the formula V = -∫E•dℓ, where E is the electric field and dℓ is the differential length along the path of integration. This means that the potential at the vertex can be found by integrating the electric field along the path from infinity to the vertex.

5. How does the potential at the vertex of a cube vary with distance?

The potential at the vertex of a cube follows an inverse-square law, meaning that it decreases as the distance from the vertex increases. This is because the electric field at the vertex also follows an inverse-square law and the potential is directly proportional to the electric field. Therefore, as the distance increases, the potential at the vertex decreases.

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