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The orginal problem statement

The electric potential from an elementary electric dipole located at the origin is given by the expression

$\phi$($\vec r$) = $\vec p$ $\cdot$ $\vec r$/4$\pi$$\epsilon_0$$r^3$

where $\vec p$ is the electric dipole moment vector. Show that the corresponding electric field is given by the expression

$\vec E$ = -$\nabla$$\phi$ = $\frac{3 (\vec p \cdot \hat r) \hat r - \vec p }{4 \pi \epsilon_0 r^3}$

where $\hat r$ is the unit vector in the direction of the vector $\vec r$.

I'm not too sure if I wrote the electric field expression correctly so I uploaded a snippet of the question which is on the attachment.

So the way I thought to solve it was by replacing $\vec r$ with $r \hat r$

so $\vec E$ = -$\frac{\partial \phi}{\partial r}\hat r$ = -$\frac{\partial}{\partial r}(\vec p$ $\cdot$ $\vec r$/4$\pi$$\epsilon_0$$r^3)\hat r$ = -$\frac{\partial}{\partial r}(\vec p$ $\cdot$ $r \hat r$/4$\pi$$\epsilon_0$$r^3)\hat r$ = $\frac{\vec p \cdot \hat r }{2\pi \epsilon_0 r^3}\hat r$

not sure what I'm doing wrong. I thought maybe since the dot product involves the angle between the two vectors one of the other components of the spherical gradient survive but I'm not sure.

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Let us use the convention $\vec{r} = x_1\hat x_1 + x_2\hat x_2+x_3\hat x_3$ and $r= |\vec{r}|$. Consider \begin{align} \psi(\vec{r}) = \vec{p}\cdot\frac{\vec{r}}{r^3} \end{align} then \begin{align} \frac{\partial}{\partial x_i} \psi(\vec{r}) = \sum_{j}p_j\cdot \frac{\partial}{\partial x_i}\left(\frac{x_j}{r^3} \right) = \sum_j p_j\cdot \frac{\delta_{ij}r^2-3x_ix_j}{r^5} = \frac{p_i}{r^3}-3\frac{x_i}{r^4}\sum_jp_j\cdot \frac{x_j}{r}. \end{align} Hence it follows \begin{align} -\nabla\psi(\vec{r}) = \frac{3(\vec{p}\cdot \hat r)}{r^3}\hat r-\frac{1}{r^3}\vec{p}. \end{align}

Note: We have use the fact that \begin{align} \frac{\partial}{\partial x_i} r = \frac{\partial}{\partial x_i}\sqrt{x_1^2+x_2^2+x_3^2}= \frac{x_i}{\sqrt{x_1^2+x_2^2+x_3^2}}=\frac{x_i}{r}. \end{align} Also, $\delta_{ij}$ is the Kronecker delta, i.e. \begin{align} \delta_{ij} = \begin{cases} 1 & \text{ if } i = j\\ 0 & \text{ otherwise} \end{cases}. \end{align}

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  • $\begingroup$ I'm not really familiar with your notation, I'm assuming that $\partial_i$ is a shorthand for $\frac{\partial}{\partial_i}$. Also, I don't understand that the expression that results from $\partial_i$$\frac{r_j}{r^3}$, what is the $\delta_(ij)$ ? $\endgroup$ – Elvis Feb 3 '17 at 6:46
  • $\begingroup$ Thank you! It seems much clearer to me now. I haven't done vector calculus in a long time so I appreciate the patience. I'm going to try to reproduce what you have done on my own and I'll post again if I get stuck. $\endgroup$ – Elvis Feb 3 '17 at 6:58
  • $\begingroup$ Ok so I've been able to follow everything, symbolically at least, up until the final result. I understand how the $p_i$ and $x_i$ got into the expression but I don't understand how you turn them into the corresponding vectors at the end. $\endgroup$ – Elvis Feb 3 '17 at 8:31
  • $\begingroup$ Never mind, I figured it out. Thanks again for the help! $\endgroup$ – Elvis Feb 3 '17 at 8:59

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