Gauss Law and Flux for variable charge density in a sphere

[engineer_ja]

Gauss Law and Flux for variable charge density in a sphere

Homework Statement

The charge density within a sphere varies as a constant, a, times its radius, r. Find an expression for the direction and magnitude of the electric flux, D, within the sphere

Homework Equations

Gauss' Law
sphere volume = [itex]\frac{4}{3}[/itex] [itex]\pi[/itex] r[itex]^{3}[/itex]

The Attempt at a Solution

I have some difficulty understanding the question, but this is what I think I should do:

Charge Density = ar

Charge in volume moving from r to r+δr , δQ, is: ar x volume
=(4/3) [itex]\pi[/itex] ar [(r+δr)^3 - r^3)]
=(4/3) [itex]\pi[/itex] ar (3 r^2 δr) (ignoring double differentials as negligible)
=4 [itex]\pi[/itex] a r^3 δr

integrating from 0 to r(an arbitrary distance from centre) gives charge within that volume, Q:
Q = [itex]\pi[/itex] a r^4

Gauss Law says flux through an area is equal to charge enclosed so;
Magnitude of the flux ,D, = Q = [itex]\pi[/itex] a r^4
Direction is radially outward (assuming charge is positive). (not sure on how to argue this...)Notes: I have no idea if this is correct, or if I'm completely of the plot! any help greatly appreciated.
I thought D was the flux density, though the question calls it flux, so do I need to divide by area?

Thanks Everyone!...

 

[KataKoniK]

Hi there,

Your approach is on the right track! Let me help you with the rest of the solution.

First, let's define the charge density, ρ, as a function of radius, r, within the sphere:

ρ(r) = ar

Now, let's consider a small element of the sphere with thickness δr and volume dV:

dV = 4πr²δr

The charge within this element is given by:

dq = ρ(r)dV = 4πar³δr

Now, we can use Gauss' Law to find the electric flux through a spherical surface of radius r:

Φ = ∫E⋅dA = Qenc / ε0

Where Qenc is the charge enclosed within the surface and ε0 is the permittivity of free space. Since the charge density is varying with radius, we need to integrate over the volume of the sphere from 0 to r:

Φ = ∫E⋅dA = ∫ρ(r)dV / ε0 = ∫0r 4πar³δr / ε0 = 4πa/ε0 ∫0r r³δr

Using the volume of a sphere formula, we can write this integral as:

Φ = 4πa/ε0 ∫0r (3/4πr³)(4πr²δr) = 3a/ε0 ∫0r r⁵δr = 3a/ε0 (r⁶/6)

Therefore, the magnitude of the electric flux through the sphere is:

|Φ| = 3a/ε0 (r⁶/6)

And the direction of the flux is radially outward, as you correctly pointed out.

I hope this helps! Let me know if you have any further questions.