Electric field of a non-conducting shell

AI Thread Summary
The discussion focuses on calculating the electric field of a hollow spherical non-conducting shell with a specific charge density. For the region inside the shell (r < a), the electric field is zero due to no enclosed charge. Between the inner and outer radii (a < r < b), the enclosed charge is derived using integration, leading to an electric field expressed as E = C(r-a)/r^2ε₀. Outside the shell (r > b), the electric field is determined similarly, resulting in E = C(b-a)/r^2ε₀. The calculations appear correct, confirming the user's understanding of the problem.
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Homework Statement



Not sure if I'm doing this problem correctly (no answer key for these practice problems). I just want to check with people that know this material well enough.

A hollow spherical non-conducting shell of inner radius a and outer radius b carries charge density p = C/r^2 in the region a =< r =< b. Find the electric field in the following regions

r < a
a < r < b
r > b

Homework Equations


[/B]
\varepsilon_0\int E \cdot dA = Qenc

The Attempt at a Solution


[/B]
for r < a

Qenc = 0 so E = 0

for a < r < b

Qenc = \int _a^r pdV

Volume of a sphere with radius r 4/3 \pi r^3

so then dV = 4\pi r^2 dr

which means Qenc = \int_a^r C/r^2 4\pi r^2 dr or \int_a^r4C \pi dr

Solving I get Qenc = 4\pi C (r-a)

Now that I have Qenc I can use

\varepsilon_0\int E \cdot dA = Qenc

using a gaussian surface of a sphere with radius r, I do

\varepsilon_0EA = 4\pi C (r-a)

A = 4\pi r^2 so that leaves me with

E = C(r-a)/r^2\varepsilon_0

for r > b

I used a similar process except I did

Qenc = \int _a^b pdV

making Qenc = 4\pi C (b-a)

so then E = C(b-a)/r^2\varepsilon_0while my answers make sense to me, I'd like to make sure I'm not making any mistakes because this question is harder than anything I've done so far!
 
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It is correct, nice work!
 
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