Gauss's Law and Non-Uniform Spherical Charge Distributions

[MrLilipili]

doctor i am very confused! is the volume of sphere not suppose to be 4/3 • π • r³...!?

could you explain why ( dQ = ρ.dv = ρ.4π • r²) ?

[usmansmir]

@MrLilipili Since the charge density is not uniform throughout the sphere's volume he needs to add up the charge in small pieces, and since the charge is a function of radius =, he comes up with a integral dependent of it: Q = integral of ρ.4π • r² • dr notice the dr is in there

[usmansmir]

@MrLilipili to make it clearer he is looking for the volume of a hollow shell with some thickness (not the sphere) which is 4π • r³(of outer thickness) - 4π • r³(of inner thickness but since the thickness of the shell is so small you can simply multiply the ""surface area"" of the shell with its thickness

[MrWonderfullable]

sur ....u are perfect ! ....i just dont understand why my teacher can't teach me this stuff simply like u do !

[ThePajiatch]

chuchuchuchucu

[2011bini]

THis is great! I must do well on my mid test =D

[Bearkiller777]

thanks that makes sense I have a test tomorrow in Electromagnetics, very helpful. might be watching a few more on a few more topics i am testing on to review.

[lasseviren1]

Yes you could. Only a few of my students have the mathematical background to do this. They are about a year away from being able to do this.

[oboeizzle]

Could you not just take a triple integral in spherical coordinates and sum up the charge density over the region? I tried this and arrived at the same answer.