Damn Gauss law is so confusing

In summary, the conversation discusses finding the radial depth beneath the outer surface of a non conducting cylindrical shell where the electric field strength is equal to one half the surface value. The method involves using Gauss' Law and a Gaussian surface to find the electric field at the outer surface of the cylinder, which is then used to find the electric field at a specific depth. A uniform volume charge density is used to calculate the charge enclosed in the Gaussian surface. The conversation also includes a discussion on how to get an explicit value for the volume charge density.
  • #1
stunner5000pt
1,461
2
Positive charge is uniformly distributed throughout a non conducting cylindrical shell of inner radius R and outer radius 2R. What radial depth beneath outer surface of the charge distribution is the elctric field strength equal to one half the surface value?

errr

e0 = epsilon 0 the permittivity

e0 EA = Q enc

thus E = Qenc / A e0

to find where E = Qenc / 2Ae0 I am totally stumped

your help would be greatly appreciated!
 
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  • #2
Start by finding the value for the electric field at the outer surface of the cylinder. We do this by making a cylindrical Gaussian surface around the cylinder. So Gauss' Law reads:

[tex]EA = \frac{Q_{enc}}{\epsilon}[/tex]

So first off we need to know what the surface area is enclosed in our Gaussian surface. The surface area of a cylinder is

[tex]A = 2\pi r l[/tex]

where [tex]l[/tex] is the length of the cylinder and [tex]r[/tex] is its radius. In our case the length is infinity but don't worry about that right now. The next step is to find the total charge that is included in our Gaussian surface. We know that the surface is UNIFORMLY distributed throughout the cylinder. We also know (obviously) that the cylinder is a three dimensional object. Therefore, we can define a uniform volume charge density [tex]\rho[/tex]that is equal to a constant.

[tex]\rho = \rho_{0} = \frac{Charge}{Volume}[/tex]

Thus, to find the amount of charge enclosed we must multiply (or integrate for a non-uniform charge density) by the volume. Thus, the charge enclosed in our Gaussian surface is

[tex]Q_{enc} = \rho_{0}V = \rho_{0} [\pi (2R)^2 - \pi R^2]l = \rho_{0}3\pi R^2l[/tex]

Thus. the electric field at the surface of the cylinder is

[tex]E(2R) =\frac{3\rho_{0}R}{2\epsilon}[/tex]

Notice that the [tex]l[/tex] in the above equations cancels out. You can get the second part in a similar way that I have shown. To get the electric field as a function of r just substitute r everywhere I had 2R above.
 
  • #3
i wouldn't have thought of using the p idea. I was wondering how to get an explicit value for p

thank you very very much!
 

Related to Damn Gauss law is so confusing

What is Gauss's law?

Gauss's law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge.

Why is Gauss's law confusing?

Gauss's law can be confusing because it involves complex mathematical concepts such as vector calculus and differential equations.

What are some applications of Gauss's law?

Gauss's law is used in many areas of physics and engineering, including electrostatics, electrodynamics, and circuit analysis. It is also used to calculate the electric field of charged objects and to understand the behavior of electric fields in various situations.

How can I better understand Gauss's law?

To better understand Gauss's law, it is important to have a strong foundation in vector calculus and to practice solving problems and applying the law to different situations. It can also be helpful to seek out additional resources, such as textbooks or online tutorials, to supplement your understanding.

Are there any common misconceptions about Gauss's law?

One common misconception about Gauss's law is that it only applies to symmetrical situations. In reality, it can be applied to a wide range of situations, including non-symmetrical ones, as long as the necessary conditions are met.

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