- #1
latentcorpse
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A sphere of radius R has a unifromly distributed charge Ze inside it (essentially it's a model of a nucleus)
now my tutor worked through the first part of the question with me and we established that the potential inside the sphere was
[itex]\varphi(\mathbf{r})=-\frac{Zer^2}{8 \pi \epsilon_0 R^3} + \frac{3Ze}{8 \pi \epsilon_0 R}[/itex] and the field inside was [itex]\mathbf{E}(\mathbf{r})=\frac{Zer}{4 \pi \epsilon_0 R^3} \mathbf{\hat{r}}[/itex]
the question I'm stuck on is showing that the electrostatic energy, [itex]\frac{1}{2} \int dV \rho \varphi[/itex], of the nucleus is equal to the field energy, [itex]\frac{1}{2} \epsilon_0 \int dV \mathbf{E}^2[/itex].
i took [itex]\rho=\frac{Ze}{\frac{4}{3} \pi R^3}=\frac{3Ze}{4 \pi R^3}[/itex]
and so electrostatic energy is
[itex]\frac{1}{2} \frac{3Z^2e^2}{32 \pi^2 \epsilon_0} \int_{r=0}^R (\frac{3}{R^4}-\frac{r^2}{R^6}) dr \int_{\theta=0}^{\pi} d \theta \int_{\phi=0}^{2 \pi} d \phi = \frac{3 Z^2 e^2}{32 \epsilon_0} [\frac{3}{R^3}-\frac{1}{3R^3}]=\frac{Z^2 e^2}{4 \epsilon_0 R^3}[/itex]
and the field energy is
[itex]\frac{1}{2} \epsilon_0 \int dV \frac{Z^2 e^2 r^2}{16 \pi^2 \epsilon_0^2 R^6}=\frac{Z^2 e^2}{32 \pi^2 \epsilon_0} \int_{r=0}^R \frac{r^2}{R^6} dr \int_{\theta=0}^{\pi} d \theta \int_{\phi=0}^{2 \pi} d \phi = \frac{Z^2 e^2}{16 \epsilon_0} [\frac{r^3}{3R^6}]_{r=0}^{R}=\frac{Z^2 e^2}{48 \epsilon_0 R^3}[/itex]
clearly I'm out by a factor of 12 but the fact that allt he constants etc are the way they should be hopefully means I'm on the right lines. can anybody help? cheers.
now my tutor worked through the first part of the question with me and we established that the potential inside the sphere was
[itex]\varphi(\mathbf{r})=-\frac{Zer^2}{8 \pi \epsilon_0 R^3} + \frac{3Ze}{8 \pi \epsilon_0 R}[/itex] and the field inside was [itex]\mathbf{E}(\mathbf{r})=\frac{Zer}{4 \pi \epsilon_0 R^3} \mathbf{\hat{r}}[/itex]
the question I'm stuck on is showing that the electrostatic energy, [itex]\frac{1}{2} \int dV \rho \varphi[/itex], of the nucleus is equal to the field energy, [itex]\frac{1}{2} \epsilon_0 \int dV \mathbf{E}^2[/itex].
i took [itex]\rho=\frac{Ze}{\frac{4}{3} \pi R^3}=\frac{3Ze}{4 \pi R^3}[/itex]
and so electrostatic energy is
[itex]\frac{1}{2} \frac{3Z^2e^2}{32 \pi^2 \epsilon_0} \int_{r=0}^R (\frac{3}{R^4}-\frac{r^2}{R^6}) dr \int_{\theta=0}^{\pi} d \theta \int_{\phi=0}^{2 \pi} d \phi = \frac{3 Z^2 e^2}{32 \epsilon_0} [\frac{3}{R^3}-\frac{1}{3R^3}]=\frac{Z^2 e^2}{4 \epsilon_0 R^3}[/itex]
and the field energy is
[itex]\frac{1}{2} \epsilon_0 \int dV \frac{Z^2 e^2 r^2}{16 \pi^2 \epsilon_0^2 R^6}=\frac{Z^2 e^2}{32 \pi^2 \epsilon_0} \int_{r=0}^R \frac{r^2}{R^6} dr \int_{\theta=0}^{\pi} d \theta \int_{\phi=0}^{2 \pi} d \phi = \frac{Z^2 e^2}{16 \epsilon_0} [\frac{r^3}{3R^6}]_{r=0}^{R}=\frac{Z^2 e^2}{48 \epsilon_0 R^3}[/itex]
clearly I'm out by a factor of 12 but the fact that allt he constants etc are the way they should be hopefully means I'm on the right lines. can anybody help? cheers.