Frequently asked questions on Gauss Law

Gauss law is a very vast topic, and there are many subtopics and formulas to it. We will mainly discuss Gauss law, what a Gaussian surface is, how to choose a Gaussian surface, etc. We will also solve some numerical problems related to Gauss law formulas.

Why does the flux ∫Ē.dĀ go to E∫dA in Gauss’s Law?

We get ∫Ē.dĀ=E∫dA when the electric field is constant over the surface you are integrating. This will be the case if you have chosen an appropriate Gaussian surface. For example, suppose the charge distribution is symmetric. In that case, a concentric Gaussian surface will have a constant electric field everywhere on the surface, and you can pull E out of the integral.

How do you choose a Gaussian surface? How do you decide what size to make it?

Generally, you pick one with the exact symmetry as the charge distribution, such that the magnitude of E is constant (or zero) over the surface. For spherical symmetry, this is a sphere: everywhere equidistant from the centre has the same E magnitude.

For cylindrical symmetry, the surface of consistent E is a cylinder. You typically pick a case with a portion of its sides corresponding to the surface for planar symmetry. For a plane corresponding to the surface, the magnitude of E is consistent. You can pick any surface size, and Gauss’s Law is valid for any size. Simply select a few sizes and give the size a name (for example, a cylinder of length L, and so forth). You will observe that this size drops toward the finish of the issue!

What is the difference between Q and q?

These labels might mean various things as per definitions in the specific problem you are doing. In the current practice issue, Q (or Q0) was the charge of the spherical ball of charge: this is the “source charge” that makes potential everywhere in space. The charge q was a little “test charge” impacted by the circle’s actual capacity.

What do Gaussian surfaces have to do with potential?

In today’s practice problem, we used Gaussian surfaces to find the electric field using Gauss’s Law. It was the first step for solving the problem via the electric field integration method. Once you know the electric field, you integrate it along a path between two points to find the potential difference between the points– that is an independent step and does not directly involve a Gaussian surface.

Suppose the electric flux of a sphere is E×4ℼr². What will be the electric field due to this flux?

Solution: The parameter is,

ɸ=E×4ℼr²

Gauss formula: ɸ=Q/E0

Therefore, E×4ℼr²=Q/E0

E= Q(4ℼr²)E0

Why do we use V = 0 at infinity?

In principle we choose V = 0. However, choosing V = 0 at infinity is very convenient for situations with spherical symmetry, including for a point charge. Then you get (for a point charge q) V = kq/r, and you do not have to drag around an extra constant.

What is Coulomb’s Law?

The rule says that the force between two charged objects corresponds to the result of the magnitude of charges and inverse relative to the square of the distance between them.

Where do Gauss’s Law and Coulomb’s Law come into play?

Gauss’s Law and Coulomb’s Law are needed in order to understand the equations that govern power and force.

When was the Gauss law discovered?

Gauss law was first discovered in 1762 by Lagrange and by Gauss in 1813.

Define Gauss law?

The flux of electric field through any closed surface, also called a Gaussian surface, is equal to the net charge enclosed divided by the permittivity of free space.

What is Q in Gauss law?

Q or q is the charge enclosed in the volume.

Can you clarify how we know that dq = λdx?

It is related to a small charge and a tiny length in a 1D problem. λ is the charge per unit length, and the charge is the length times the charge per length. Here is a way to think about a uniform (constant) lambda: the ratio of total charge Q to the full-length L is Q L. Now, if you take any part of the length ∆x, the ratio of the charge ∆Q in that length to ∆x is ∆q ∆x = Q L = λ. Taking infinitesimals, we have dq = λdx. This relation still works for a non-constant λ, so long as you are considering charge elements small enough that there is negligible variation between them.

Conclusion

Discussing all these significant points of Gauss Law under the topic Frequently asked question has cleared the doubt regarding many common questions that are frequently asked and yet not known to many people. We discussed Gaussian surfaces, potential inside a shell, electric potential energy, and electric flux.