Consider a constant electric dipole moment \(\mathrm{p}\) at the origin. It produces an electric potential of \(\phi=\frac{\mathrm{p} \cdot \mathrm{r}}{4 \pi \epsilon_{0} r^{3}}\) outside the dipole. Noting that \(\mathbf{E}=-\nabla \phi\), find the electric field at \(\mathbf{r}\).

The electric potential \( \phi \) outside the dipole is given as \( \phi=\frac{\mathrm{p} \cdot \mathrm{r}}{4 \pi \epsilon_{0} r^{3}} \). This equation is in dot product form, hence it's a scalar equation. Convert it into a vector form to facilitate the application of the gradient operator. The scalar notation \(\mathrm{p} \cdot \mathrm{r}\) denotes the dot product of \(\mathrm{p}\) and \(\mathrm{r}\). This can be translated to vector notation as \( |\mathrm{p}| |\mathrm{r}|cos(\theta) \) where |x| denotes the magnitude of a vector x, and \( \theta \) is the angle between vectors p and r. Hence, the electric potential \( \phi \) can be expressed as \( \phi = \frac {|\mathrm{p}| |\mathrm{r}|cos(\theta)} {4 \pi \epsilon_{0} r^{3}} \).

To find the electric field \(\mathbf{E}\), we need to apply the gradient operator to the electric potential. The gradient (or del, denoted as \( \nabla \)) in spherical polar coordinates is given as \( \nabla \phi = \frac {\partial \phi}{\partial r} \hat{r} + \frac {1}{r} \frac {\partial \phi}{\partial \theta} \hat{\theta} \). Here, \( \hat{r} \) and \( \hat{\theta} \) are the unit vectors in the radial and angular directions respectively. Apply the gradient operator to \( \phi \) to derive the expression for the electric field \( \mathbf{E} \).

The negative of the gradient provides the electric field \( \mathbf{E} \). Hence, by applying the negative operator, we get \( \mathbf{E} = - \nabla \phi = - \frac {\partial \phi}{\partial r} \hat{r} - \frac {1}{r} \frac {\partial \phi}{\partial \theta} \hat{\theta} \). This is the electric field at \( \mathbf{r} \) in vector form.