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My question is abut finding components of electric field from potential function. $V=\frac{1}{4\pi\epsilon_0}\frac{\vec{p}\cdot\vec{r}}{r^3}$ (1)

I can write it as $V=\frac{1}{4\pi\epsilon_0}\frac{pz}{r^3}$ (2)

By taking partial derivative

$\vec{E_x}=\frac{\partial{V}}{\partial{x}}$

$\vec{E_x}=\frac{pz}{4\pi\epsilon_0}\frac{\partial\frac{1}{r^3}}{\partial{x}}$

$\vec{E_x}=\frac{pzx\hat{x}}{4\pi\epsilon_0r^3}$

So is $\vec{E_y}$

$\vec{E_y}=\frac{pzx\hat{y}}{4\pi\epsilon_0r^3}$

However

$\vec{E_z}=\frac{p}{4\pi\epsilon_0}\frac{\partial{\frac{z}{r^3}}}{\partial{z}}$

$\vec{E_z}=\frac{p}{4\pi\epsilon_0}\left[\frac{3z^3}{r^3}-\frac{1}{r^3}\right]$

Here I get $\vec{E}=\frac{1}{4\pi\epsilon_0r^3}\left(3pz\vec{r}-p\hat{z}\right)$

But i am supposed to find

$\vec{E}=\frac{1}{4\pi\epsilon_0r^3}\left(\frac{3\left(\vec{p}\cdot\hat{r}\right)}{r^2}-\vec{p}\right)$

What is my mistake here? I tried but could not see. By the war from (1) to (2) $\vec{p}\cdot\vec{r}=pr\cos{\alpha}$

$\cos{\alpha}=\frac{z}{r}$

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closed as off-topic by ACuriousMind, Bill N, John Rennie, Danu, Kyle Kanos Mar 23 '16 at 10:06

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  • $\begingroup$ Right before i wrote" Here i get" that 3z^3 is not the case but 3z^2 instead? $\endgroup$ – user96369 Mar 22 '16 at 22:52
  • $\begingroup$ I would recommend using the spherical gradient instead of the Cartesian one: $\nabla =\hat{r} \frac{\partial}{\partial r} + \hat{\theta} \frac{1}{r} \frac{\partial}{\partial \theta} + \hat{\phi} \frac{1}{r \sin{\theta}} \frac{\partial}{\partial \phi}$ $\endgroup$ – JoDraX Mar 22 '16 at 23:14
  • $\begingroup$ There's something wrong with your "supposed to find" expression, because in the parentheses, you have a vector subtracted from a scalar. $\endgroup$ – Brionius Mar 22 '16 at 23:28
  • $\begingroup$ Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Mar 23 '16 at 6:20