PowerWill
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Working on Griffiths 3.33. I'm supposed to show that the Electric Field of a pure dipole can be written in the following coordinate free form:
[tex] \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0 r^3} [3(\vec{p} \cdot \hat{r})\hat{r} - \vec{p}][/tex]
Where p is the dipole. I know that the potential is equal to
[tex] V(r,\theta) = \frac{\hat{r} \cdot \vec{p}}{4 \pi \epsilon_0 r^2}[/tex]
and I tried to take the negative gradient of that, but got lost in the math. If you assume the dipole points along the z-axis you get the solution
[itex] \vec{E}(r,\theta) = \frac{p(2cos \theta \hat{r} + sin \theta \hat{\theta})}{4 \pi \epsilon_0 r^3}[/itex]
And I tried to work with that a little to no avail. Any ideas how to solve this beast?
[tex] \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0 r^3} [3(\vec{p} \cdot \hat{r})\hat{r} - \vec{p}][/tex]
Where p is the dipole. I know that the potential is equal to
[tex] V(r,\theta) = \frac{\hat{r} \cdot \vec{p}}{4 \pi \epsilon_0 r^2}[/tex]
and I tried to take the negative gradient of that, but got lost in the math. If you assume the dipole points along the z-axis you get the solution
[itex] \vec{E}(r,\theta) = \frac{p(2cos \theta \hat{r} + sin \theta \hat{\theta})}{4 \pi \epsilon_0 r^3}[/itex]
And I tried to work with that a little to no avail. Any ideas how to solve this beast?