Solving Gauss's Law Problem for a Solid Cylinder

In summary, an infinitely long solid cylinder with radius R1 and volume charge density A.r is considered. The amount of charge enclosed by a cylindrical Gaussian surface with length L and radius r, concentric with the x axis, is derived for r greater than or equal to R1 and for r less than or equal to R1. Using Gauss's Law, an expression for the electric field as a function of r is found for these two regions. A graph of the magnitude of the electric field is also shown. Another method using the differential form of Gauss's Law is also mentioned.
  • #1
purplex76
6
0
An infinitely long solid cylinder radius R1 lies with it's central cylindrical axis lying along the x axis. it is made of a non-conducting material. It has a volume charge density that varies with readius as follows... p(r)=A.r (C/m^3)
where A is a constant. Consider a cylindrical Gaussian surface of length L, radius r, concentric with the x axis.

1) Derive a formula for the amount of charge enclosed by this Gaussian surface for r is greater than or equal to R1, and for r is less than or equal to R1

2) Use gauss's Law to find an expression for the electric field as a function of r in these two regions

3) graph the magnitude of the electric filed for these two regions.


i would appreciate any help with this question because it is really stumping me...Thanks!
 
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  • #2
What exactly have you tried so far and what is your thinking?
 
  • #3
Oh wait one more thing, please post this type of question under "college level help" thank you.
 
  • #4
Solution to the Gauss law problem:

Volume charge density= Ar
a -> radius of cylinder
For r > a

Let the radius of cylindrical Gaussian surface be r

E . 2. pi. r. l = integral {( 2*pi.A.l.r. dr / e0 ), 0 , a}

[integral { (), ,} denotes-- () - integral funciton then the limits]

e0 -> permittivity of free space

[ E = A. (a^2)/ (2*r) ] ... Solution

For r<=a

E . 2*pi.r.l = integral { (2*pi.A.l.r dr/ e0), 0, r}


[ E=A.r/2 ] ..... Solution
 
  • #5
that doesn't make sense to me. Isn't the problem more complex than 2 integrals, because i got no credit for the integrals i put down, being somewhat similar to the ones you replied with.
 
  • #6
gauss law problem

:frown: [tex]\mbox{i m sorry i forgot to divide by } \epsilon_0\mbox{. Divide the solutions by} \epsilon_0 \mbox{. I feel, that is the correct solution.}[/tex]
 
  • #7
another method

You may also use the differential form of Gauss law for cylindrically radial field. It goes something like this:
[tex] \frac{d(E.r)}{dr} = \frac{\rho r}{\epsilon_0} [/tex]

Make [tex] \rho [/tex] as a function or r and integrate over proper limits.
 

1. What is Gauss's Law and why is it important in solving problems for a solid cylinder?

Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. In the case of a solid cylinder, it allows us to determine the electric field at any point on the surface or inside the cylinder by calculating the charge enclosed by a cylindrical surface.

2. What are the steps for solving Gauss's Law problem for a solid cylinder?

The steps for solving Gauss's Law problem for a solid cylinder are:

1. Choose a cylindrical Gaussian surface that encloses the cylinder.

2. Calculate the charge enclosed by the Gaussian surface.

3. Apply Gauss's Law, which states that the electric flux through the surface is equal to the charge enclosed divided by the permittivity of free space.

4. Use the symmetry of the cylinder to simplify the calculation of electric flux.

5. Solve for the electric field at any point on the surface or inside the cylinder.

3. What assumptions are made when solving Gauss's Law problem for a solid cylinder?

The main assumptions made when solving Gauss's Law problem for a solid cylinder are:

1. The cylinder has a uniform charge distribution.

2. The cylinder is infinitely long, so the ends can be neglected in the calculation.

3. The electric field is perpendicular to the surface of the cylinder.

4. Can Gauss's Law be used to solve problems for a hollow cylinder?

Yes, Gauss's Law can also be applied to solve problems for a hollow cylinder. However, the shape of the Gaussian surface and the calculation of charge enclosed will be different compared to a solid cylinder.

5. Are there any real-life applications of solving Gauss's Law problem for a solid cylinder?

Yes, there are many real-life applications of solving Gauss's Law problem for a solid cylinder. For example, it can be used to determine the electric field inside a charged wire, which is important in understanding and designing electrical circuits. It is also used in the study of electrostatic discharge and lightning protection systems.

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