# Gauss Law Problem with Cylinders

• physics16102
In summary, the electric field at a distance of 0 from the axis of an infinitely long, non-conducting cylinder of radius R is given by E(2πRL)=(ρπr^2L)/εo.
physics16102

## Homework Statement

An infinitely long, non-conducting cylinder of radius R carries a uniform volume charge density ρ. Find the magnitude of the electric field for 0 < r < R

EA=Qin/εo

## The Attempt at a Solution

I am debating whether the answer should be ρr^2/2εoR or ρr/2εo.
I said E(2πRL)=(ρπr^2L)/εo and solved to get ρr^2/2εoR.

I was confident in my answer but then found the same problem in the book, but it asked to find E at distance r from axis, where r<R. The answer to this was ρr/2εo. Can someone explain the difference between these two problems, if there is one. They may be exactly the same and my calculation is wrong, but I am confused over the two

physics16102 said:
I said E(2πRL)=(ρπr^2L)/εo and solved to get ρr^2/2εoR.

If you want to find E at r<R your Gaussian surface is a cylinder of radius r.

ehild

Since r<R, R will never appear in the equation since my dA vector relates to r and Qin also relates to r. I would simply replace R with r in my equation and solve, yielding ρr/2εo. Is this logic correct?

I think I just failed to remember that in the basic form of the equation, you are integrating over the Gaussian surface. But does Qin relate charge enclosed by the Gaussian volume drawn or by the entire volume in question?

Qin is the charge of volume enclosed by the Gaussian surface.

ehild

physics16102 said:
So I would get E(2πrL)=(ρπr^2L)/εo and E=ρr/2εo.
But this website says I am wrong so I am now very confused
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html#c3

It is saying for r<R, E should be completely different. Can someone please clear this up?

In that formula at hyperphysics, E is given in terms of lambda, the charge per unit length. In your problem, the charge per unit volume (rho)was given. Compare your new result with that you cited from your book in your first post.

ehild

When I calculate it, I get ρr/2εo.
But the website above says something different, so I am clueless as to what to put if this problem were on my test tomorrow(today lol).

We posted at the same time. I did not notice that. I'll change my calculation and see what I get

ehild

I've concluded that the calculation is correct for my problem and it is ρr/2εo.
But, I don't understand where they get the R^2 in the bottom of their equation on hyperphysics. It has to come from the A part of EA=Qin/eps, but how? Shouldn't A always be in terms of the gaussian surface, which has nothing to do with R.

nevermind, found the link that explains it. I would never come up with that on my own but at least I now understand the difference and why it is there

## 1. What is Gauss Law?

Gauss Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is named after the German mathematician and physicist, Carl Friedrich Gauss.

## 2. How does Gauss Law apply to cylinders?

Gauss Law can be applied to cylinders by considering the cylinder as a closed surface and calculating the electric flux through it. This can be useful in determining the electric field at any point within or outside the cylinder.

## 3. What are the key equations involved in solving Gauss Law problems with cylinders?

The key equations involved in solving Gauss Law problems with cylinders are the electric flux equation, which relates the electric flux through a closed surface to the enclosed charge, and the electric field equation, which relates the electric field to the charge distribution within the cylinder.

## 4. How do you choose the Gaussian surface for a cylinder in a Gauss Law problem?

The Gaussian surface for a cylinder should be chosen such that it encloses the charge or charges you are interested in finding the electric field for. This surface should also be chosen to have a high degree of symmetry to simplify the calculations.

## 5. What are some common mistakes made when solving Gauss Law problems with cylinders?

Some common mistakes made when solving Gauss Law problems with cylinders include not properly choosing the Gaussian surface, not considering the direction of the electric field, and not taking into account the symmetry of the problem. It is important to carefully consider these factors in order to accurately solve the problem.

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