Thick spherical shell confirmation of solution

[Liquidxlax]

Homework Statement

A spherical shell of inner radius a and outer radius b contains a uniform charge density ρ of total charge Q. (a) Find ρ (b) find the electric field for r< a, a< r< b, b< r sketch it

Homework Equations

E = kQ/r²

V=4πr³

The Attempt at a Solution

ρ π

since it is a thick shell the volume would be V = (4/3)π(b³ - a³)

and ρV = Q_enc => ρ=Q/V

so

ρ = 3Q/(4π(b³ - a³))


(b) r<a the electric field will be 0 because the electric field can't close on itself

a< r< b

E=kQ_enc/r² = (4πkρ(b³ - a³))/3r³


b<r kQ/r² because the sphere can be treated as a point charge and the electric field is symmetric on a sphere


Does this look right?

 

[collinsmark]

E= [...] = (4πkρ(b³ - a³))/3r³

There's two errors with the above equation (see the red).

 

[Liquidxlax]

There's two errors with the above equation (see the red).

okay i see how the r³ is wrong, that was just a transcribing error, but how come the b is wrong? or would it be because we know that r<b so we can assume it to be

(r³-a³) because a is constant and r does not exceed b

 

[collinsmark]

There you go! Remember Gauss' law relates to the charge enclosed within the Gaussian surface (the Gaussian surface in this case has radius r). In the case where b > r > a, then any charge between r and b is outside the Gaussian surface, and has no effect (in this spherically symmetrical case).

 

[rwisz]

Your attempt at a solution looks mostly correct. However, there are a few things that could be clarified or expanded upon:

1. In part (a), it would be helpful to explicitly state that Qenc = Q, since the shell contains all of the charge Q.

2. In part (b), you could also mention that the electric field inside the shell (a < r < b) is 0 because the charges within the shell cancel each other out, resulting in no net electric field.

3. For the electric field outside the shell (r > b), it would be more accurate to say that the electric field is the same as that of a point charge with charge Q located at the center of the sphere. This is because the charges within the shell can be treated as a point charge at the center of the shell, and the electric field due to this point charge is the same as the electric field outside the shell.

Overall, your solution demonstrates a good understanding of the problem and the relevant equations. Well done!