PowerWill
- 9
- 0
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:
<br /> \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0 r^3} [3(\vec{p} \cdot \hat{r})\hat{r} - \vec{p}]<br />
Where p is the dipole. I know that the potential is equal to
<br /> V(r,\theta) = \frac{\hat{r} \cdot \vec{p}}{4 \pi \epsilon_0 r^2}<br />
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
<br /> \vec{E}(r,\theta) = \frac{p(2cos \theta \hat{r} + sin \theta \hat{\theta})}{4 \pi \epsilon_0 r^3}<br />
And I tried to work with that a little to no avail. Any ideas how to solve this beast?
<br /> \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0 r^3} [3(\vec{p} \cdot \hat{r})\hat{r} - \vec{p}]<br />
Where p is the dipole. I know that the potential is equal to
<br /> V(r,\theta) = \frac{\hat{r} \cdot \vec{p}}{4 \pi \epsilon_0 r^2}<br />
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
<br /> \vec{E}(r,\theta) = \frac{p(2cos \theta \hat{r} + sin \theta \hat{\theta})}{4 \pi \epsilon_0 r^3}<br />
And I tried to work with that a little to no avail. Any ideas how to solve this beast?