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I have 3 components, $r$, $\theta$ and $\phi$, for an electric field in spherical coordinates (and the $\phi$ component happens to be zero), let's say I just want to convert the $r$ component into cartesian, which looks like:

$$ -\frac{0.058125 \cos\theta\sin^2\theta}{r^3} $$

How do I convert this into cartesian?

Edit: Sorry maybe I should have explained that this expression is one component of a vector, which I got using E= −∇V

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  • $\begingroup$ See this: en.wikipedia.org/wiki/Gradient $\endgroup$ – lucas Apr 19 '16 at 6:53
  • $\begingroup$ don't forget about the unit vector - maybe you intend to transform that too. $\endgroup$ – anon01 Apr 19 '16 at 14:19
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You said this is the $r$-component, then you've missed the $\hat{\bf r}$.

Use $$r=\sqrt{x^2+y^2+z^2}$$ $$\cos\theta=z/r=\frac{z}{\sqrt{x^2+y^2+z^2}}$$ $$\hat{\bf r}=\frac{x\hat{\bf x}+y\hat{\bf y}+z\hat{\bf z}}{r}$$

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  • $\begingroup$ what if I had a sin(phi) or cos(phi) in the expression? $\endgroup$ – user43783 Apr 19 '16 at 7:49
  • $\begingroup$ Use $\tan\phi=y/x$. $\endgroup$ – velut luna Apr 19 '16 at 8:07
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Assuming that $\theta$ is the polar angle (angle between $\vec r$ and $\hat z$) and $\phi$ the azimuthal angle then the following relationships can be used.

$x = r \;\sin \theta \;\cos \phi$
$y = r \;\sin \theta\;\sin \phi$
$z= r\; \cos \theta$

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  • $\begingroup$ these aren't very helpful as written - you need the inverted relations $\endgroup$ – anon01 Apr 19 '16 at 14:17

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