- #1
WraithM
- 32
- 0
I don't really know much about serious business electrostatics. I've only taken the AP Physics C E&M exam (using the integral form), but I was looking at wikipedia and I was curious about the differential form of Gauss' Law.
I don't understand what I'm doing wrong with this. I'm trying to derive coulomb's law (just so I get a better understanding of how to use the differential form) from the differential form of Gauss' law.
So, you have
[tex]\rho (r) = \frac{Q}{V(r)}[/tex]
and I guess for a point charge, you'd use a spherical volume, so,
[tex]V(r) = \frac{4}{3}\pi r^3[/tex]
Therefore,
[tex]\rho (r) = \frac{3Q}{4\pi r^3}[/tex]
Then plug that into Gauss' law:
[tex]\nabla \cdot E = \frac{3Q}{4\pi \epsilon_0 r^3}[/tex]
Because there's spherical symmetry, I'm just going to ignore the rest of the divergence operator and just say
[tex]\nabla \cdot E = \frac{dE}{dr}[/tex]
Perhaps this is where I went wrong, and the divergence is way more complicated than that for this situation, but this makes sense to me intuitively.
Then you'd have after separating variables and ignoring a negative sign:
[tex]E = \int_\infty^r \frac{3Q}{4\pi\epsilon_0 r^3} dr[/tex]
[tex]E = \frac{3Q}{8\pi\epsilon_0 r^2}[/tex]
[tex]\frac{3Q}{8\pi\epsilon_0 r^2} \neq \frac{Q}{4\pi\epsilon_0 r^2}[/tex]
lol wat? There's a 2 in the denominator when there should be a 3 to make the 3 in the top cancel out and be happy. What am I doing wrong? Or am I just making utter non-sense?
I don't understand what I'm doing wrong with this. I'm trying to derive coulomb's law (just so I get a better understanding of how to use the differential form) from the differential form of Gauss' law.
So, you have
[tex]\rho (r) = \frac{Q}{V(r)}[/tex]
and I guess for a point charge, you'd use a spherical volume, so,
[tex]V(r) = \frac{4}{3}\pi r^3[/tex]
Therefore,
[tex]\rho (r) = \frac{3Q}{4\pi r^3}[/tex]
Then plug that into Gauss' law:
[tex]\nabla \cdot E = \frac{3Q}{4\pi \epsilon_0 r^3}[/tex]
Because there's spherical symmetry, I'm just going to ignore the rest of the divergence operator and just say
[tex]\nabla \cdot E = \frac{dE}{dr}[/tex]
Perhaps this is where I went wrong, and the divergence is way more complicated than that for this situation, but this makes sense to me intuitively.
Then you'd have after separating variables and ignoring a negative sign:
[tex]E = \int_\infty^r \frac{3Q}{4\pi\epsilon_0 r^3} dr[/tex]
[tex]E = \frac{3Q}{8\pi\epsilon_0 r^2}[/tex]
[tex]\frac{3Q}{8\pi\epsilon_0 r^2} \neq \frac{Q}{4\pi\epsilon_0 r^2}[/tex]
lol wat? There's a 2 in the denominator when there should be a 3 to make the 3 in the top cancel out and be happy. What am I doing wrong? Or am I just making utter non-sense?