Question

A Uniform charge density of \(\color{blue}{2 \; \; C/m^3}\) exists in the volume \(\color{green}{2 \le x \le 4}\). Use Gauss's Law to find \(\color{red}{\textbf{D}}\) in all regions..

Answer 1

@aum , I need to close this, as I have another question to post, please when you come, do give me guidance here, how to proceed in this..

I have doubt here, as For making volume, all the three ie x, y and z should be variable, here only x is varying, what will I assume for y and z ??

Answer 2

The uniform charge density is in the region:

\(2 \le x \le 4; ~~~-\infty \lt y \lt \infty; ~~~ -\infty \lt z \lt \infty.\)

Due to symmetry, D will exist only in the x-direction. It will cancel out in the y and z directions. Choose a suitable Gaussian surface and consider three cases when computing D:
Case 1: What is D when x < 2?
Case 2: What is D when x > 4?
Case 3: What is D when \(2 \le x \le 4\) ?

Answer 3

Oh like this:

For x < 2, As no charge is enclosed, so D = 0

For x > 4 also, D = 0

For \(2 \le x \le 4\), \(Q_{enclosed} = \int \rho \cdot dv\)

And here, I am stuck..

If I choose Cylindrical Gaussian Surface, then Enclosed Charge will be Infinity as it is extending upto infinity..

And I am going very wrong I think..:(

Answer 4

Since the problem is in Cartesian coordinates and the charge distribution is in the form of a slab that is 2m wide along the x-axis and infinitely long in the y and z directions, a cuboid may be more appropriate for the Gaussian surface: