What Is the Electric Field Inside a Sphere with Varying Charge Density?

[shinobi20]

Homework Statement

A sphere with radius R has a spherically symmetric charge density that varies as 1/r. What is the electric field outside and inside the sphere?

Homework Equations

E=kQ/r^2, ε=permitivity of free space, Q=total charge, ρ=charge density, dτ=infinitesimal volume

The Attempt at a Solution

For the case (outside), due to the concept that we can treat the sphere as a point charge, E = (1/4*pi*e)(Q/r^2).

For the case (inside), by using a Gaussian surface, we have E (4πr^2) = Q/ε. By evaluating Q = ∫ ρ dτ. I got E = (R^2)/(2εr^2).

Is this correct? Or am I missing something?

 

[BvU]

I don't see them matching up at r = R

Your inside case solution description isn't detailed enough for me to tell you where you derail.
And your problem description has holes in it. Is Q the total charge on the sphere? Is it a given ? What does that mean for the enclosed charge when r < R ?

 

[shinobi20]

Sorry, there was a typo, it should be E(4πr^2) = Q/ε and Q is the total charge. What I've done for the inside is, given ∫E⋅da = Q/ε, we have E(4πr^2) = Q/ε. But Q=∫ρ dτ and ρ=1/r. So Q=∫ dτ/r. Also, dτ= r^2 sinθ dr dθ d∅ is the change of the volume in spherical coordinates. So, Q=∫ r sinθ dr dθ d∅. 0≤r≤R, 0≤θ≤π, 0≤∅≤ 2π. By integration, I got Q = (4πR^2)/2. Thus, I got E = (R^2)/ 2εr^2.

 

[BvU]

Aha! Check your integration bounds for the contained charge when r < R ! Contained charge as a function of r is not a constant ! You confuse yourself by using the symbol Q both for the total charge and also for contained charge as a function of r...

 

[shinobi20]

Oh! So to rephrase, E(4 π r'^2) = Qenc/ε with Qenc as the enclosed charge in the gaussian surface. Qenc= ∫ r sinθ dr dθ d∅. 0≤r≤r', 0≤θ≤π, 0≤∅≤ 2π. Qenc = (4 π r'^2)/2. But this will cancel the r'. E(4 π r'^2) = (4 π r'^2)/2ε.

 

[BvU]

So nice it doesn't diverge for r to 0, and that now it matches at r = R !

 

[shinobi20]

Aha! Check your integration bounds for the contained charge when r < R ! Contained charge as a function of r is not a constant ! You confuse yourself by using the symbol Q both for the total charge and also for contained charge as a function of r...

I've done everything over and over, If I integrate, the two r' really cancel, is there something wrong here? I'm confused.

 

[BvU]

I think it's the right answer.

 

[shinobi20]

I think it's the right answer.

What do you mean right? But that will mean the electric field is not varying as we go from r=0 to r=r'. How is that?

 

[BvU]

It means area and charge contained grow with the same power of r . In another example (number 4 here) charge grows with r³ and area with r², so E grows with r¹

 

[shinobi20]

Thank you so much! That is kinda subtle.