# Using a cube as a Gaussian surface

• anban
In summary, a classmate told me that it would be very difficult to use a Gaussian surface to compute the value of the E field at some arbitrary point outside the charged cube. However, if I simplify the problem to just be a point charge surrounded by a cubical Gaussian surface, then the E field calculation becomes much more simple.
anban

## Homework Statement

Suppose there is a uniformly charged cube with known side length. I then imagine a larger, closed cube surface surrounding it. This larger cube has double the side length and is symmetrical to the smaller cube.

Is is practical to use this Gaussian surface to compute the value of the E field at some arbitrary point outside the charged cube?

## Homework Equations

Gauss' Law, $\int$E*da = q/$\epsilon$

## The Attempt at a Solution

A classmate told me that this would be very difficult to do, and that no, using a cubical Gaussian surface would not work very well in a practical sense.

When I think of this problem, this is what I think:
Every charge on the surface of the inner cube will have an electric field pointing away from its center radially. Since the cube is symmetrical, though, many of these field lines will cancel each other out, leaving only field lines going outward from the inner cube perpendicular to each cube face. That would mean that the larger, imaginary cube would experience a flux out of each of its cube faces equal to the E field coming from each inner cube face. The sum of each face would then allow us to know the total flux leaving the Gaussian surface.

However, this would give us a discontinuous E field, one that would have no flux or E field coming off the edges. This would give us E field lines in a 3D cross shape.

I have no clue if this problem is very simple or very difficult. Any insights are very warmly welcome.

One more idea for solving this:

If I simplify the the inner charged cube to just be a point charge surrounded by a cubical Gaussian surface, then computing the E field is very difficult. It is not simple because the electric field vector coming from the charge is often at an angle to the area vector of the imaginary surface. It would be a very difficult integral to execute.

I don't see how Gauss can help here.

I would orient the cube with its center at (0,0,0) and observation point along the x axis, then compute the potential by integration and finally take the negative gradient of that potential. At least that way you only have to integrate over 1/4th the cube volume.

## 1. How does using a cube as a Gaussian surface simplify calculations?

Using a cube as a Gaussian surface allows for symmetrical calculations, as all sides of the cube have equal area and the electric field is constant on each side. This simplifies the integration process and makes it easier to calculate the electric flux through the surface.

## 2. Can any shape be used as a Gaussian surface?

No, only symmetrical shapes such as spheres, cubes, and cylinders can be used as Gaussian surfaces. This is because the electric field must be constant on each side of the surface in order to simplify the calculations.

## 3. How do you determine the orientation of the Gaussian surface?

The orientation of the Gaussian surface is determined by the direction of the electric field. The surface should be perpendicular to the electric field lines in order to maximize the flux through the surface.

## 4. Is the electric flux through a cube Gaussian surface always the same?

No, the electric flux through a cube Gaussian surface can vary depending on the strength and direction of the electric field. However, it will always be constant on each side of the cube as long as the electric field is constant.

## 5. Can a cube Gaussian surface be used to calculate the electric field at a specific point?

Yes, by using Gauss's law and knowing the charge enclosed within the cube Gaussian surface, the electric field can be calculated at a specific point outside the surface. However, a different method must be used to calculate the field at a point inside the surface.

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