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1. Homework Statement
(Note: Don't worry about significant digits. I just want to be able to do the question and will worry about significant digits on the exam.)
In the figure below, a nonconducting spherical shell of inner radius a=2.00cm and outer radius b=2.40cm has (within its thickness) a positive volume density [tex]\rho[/tex]=A/r, where A is a constant and r is the distance from the centre of the shell. In addition, a small ball of charge q=45.0 fC is located at the centre.
What value should A have if the electric field in the shell (a[tex]\leq[/tex]r[tex]\leq[/tex]b) is to be uniform?
(See attached file for the picture. sorry it isn't very good. That's a "q" beside the +ve charge in the middle.)
2. Homework Equations
Charge density : [tex]\rho= \frac{dq}{dV}[/tex]
Volume : [tex]V = 4\pi r^3[/tex]
Electric field : [tex]E = \frac{1}{4\pi \epsilon} \int \frac{dq}{r^2}[/tex]
3. The Attempt at a Solution
E at a distance r anywhere within [a, b] is constant.
E is caused by the point charge and distribution of charges in the volume.
[tex]E_{net}[/tex] is constant, therefore
[tex]E_{charge}=E_{volume}[/tex]
[tex] \frac{1}{4\pi \epsilon} \int \frac{dq_{charge}}{r_{charge}^2} =  \frac{1}{4\pi \epsilon} \int \frac{dq_{volume}}{r_{volume}^2}[/tex]
the E for a point charge doesn't need to have an integral, because it works out that way.
[tex] \frac{q_{charge}}{r^2} =  \int \frac{dq_{volume}}{r_{volume}^2}[/tex]
Charge density : [tex]\rho= \frac{A}{r} = \frac{dq}{dV}[/tex]
[tex]dq = \frac{A}{r} dV[/tex]
[tex] \frac{q_{charge}}{r^2} =  \int \frac{A dV}{r^3}[/tex]
dV is related to r because [tex]V = \frac{4}{3} \pi (r^3  a^3)[/tex]
then [tex]r^3 = \frac{3V + 4 \pi a^3}{4 \pi}[/tex]
[tex] \frac{q_{charge}}{r^2} =  \int \frac{4 \pi A dV}{3V + 4 \pi a^3}[/tex]
[tex] \frac{q_{charge}}{r^2} =  4 \pi A \int \frac{dV}{3V + 4 \pi a^3}[/tex]
[tex] \frac{q_{charge}}{r^2} =  \frac{4 \pi A}{3} ln (3V + 4 \pi a^3)[/tex]
I don't know what to do after this, and I don't know what to do. I don't know if I messed up something.
This may be supposed to be done with Gauss' Law, but I couldn't figure out what to do with that.
Gauss' Law : [tex]\Phi = \oint E da[/tex] (In this case since E and a are in the same direction, the dot product just became E da)
(Note: Don't worry about significant digits. I just want to be able to do the question and will worry about significant digits on the exam.)
In the figure below, a nonconducting spherical shell of inner radius a=2.00cm and outer radius b=2.40cm has (within its thickness) a positive volume density [tex]\rho[/tex]=A/r, where A is a constant and r is the distance from the centre of the shell. In addition, a small ball of charge q=45.0 fC is located at the centre.
What value should A have if the electric field in the shell (a[tex]\leq[/tex]r[tex]\leq[/tex]b) is to be uniform?
(See attached file for the picture. sorry it isn't very good. That's a "q" beside the +ve charge in the middle.)
2. Homework Equations
Charge density : [tex]\rho= \frac{dq}{dV}[/tex]
Volume : [tex]V = 4\pi r^3[/tex]
Electric field : [tex]E = \frac{1}{4\pi \epsilon} \int \frac{dq}{r^2}[/tex]
3. The Attempt at a Solution
E at a distance r anywhere within [a, b] is constant.
E is caused by the point charge and distribution of charges in the volume.
[tex]E_{net}[/tex] is constant, therefore
[tex]E_{charge}=E_{volume}[/tex]
[tex] \frac{1}{4\pi \epsilon} \int \frac{dq_{charge}}{r_{charge}^2} =  \frac{1}{4\pi \epsilon} \int \frac{dq_{volume}}{r_{volume}^2}[/tex]
the E for a point charge doesn't need to have an integral, because it works out that way.
[tex] \frac{q_{charge}}{r^2} =  \int \frac{dq_{volume}}{r_{volume}^2}[/tex]
Charge density : [tex]\rho= \frac{A}{r} = \frac{dq}{dV}[/tex]
[tex]dq = \frac{A}{r} dV[/tex]
[tex] \frac{q_{charge}}{r^2} =  \int \frac{A dV}{r^3}[/tex]
dV is related to r because [tex]V = \frac{4}{3} \pi (r^3  a^3)[/tex]
then [tex]r^3 = \frac{3V + 4 \pi a^3}{4 \pi}[/tex]
[tex] \frac{q_{charge}}{r^2} =  \int \frac{4 \pi A dV}{3V + 4 \pi a^3}[/tex]
[tex] \frac{q_{charge}}{r^2} =  4 \pi A \int \frac{dV}{3V + 4 \pi a^3}[/tex]
[tex] \frac{q_{charge}}{r^2} =  \frac{4 \pi A}{3} ln (3V + 4 \pi a^3)[/tex]
I don't know what to do after this, and I don't know what to do. I don't know if I messed up something.
This may be supposed to be done with Gauss' Law, but I couldn't figure out what to do with that.
Gauss' Law : [tex]\Phi = \oint E da[/tex] (In this case since E and a are in the same direction, the dot product just became E da)
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