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Trying to work through this question I am sure that I am doing something wrong.

We start with a uniform electric field

$$ E = E_o \hat{i} $$

and then find that the potential is given by

$$ V=-E_0 x + A$$

where A is a constant In cylindrical coordinates this gives

$$ V(r,\phi,z)= -E_0 r \cos(\phi) $$

And hence the electric field in cylindrical coordinates is given by

$$ -\nabla V = -\left({\partial V \over \partial r}\hat{r} + {1\over r}{\partial V \over \partial \phi}\hat{\phi}+ {\partial V \over \partial z}\hat{z}\right) \\= +E_0 \cos(\phi) \hat{r}-E_0 {1\over r}r\sin(\phi)\hat{\phi} \\ = E_0 (\cos(\phi)\hat{r} -\sin(\phi)\hat{\phi})$$

Now my confusion is in the following step where I try to calculate the magnitude of the electric field in the cylindrical coordinate system.... .... I can see how to fix it so that I get the required answers of $E_0$, but to my mind to calculate the magnitude of the vector in space I should be taking the root sum square of three components; the component in $r$, the component in $\phi$ multiplied by $r$ and the component in $z$... if we were to calculate a volume we would use $dr \cdot rd\phi \cdot dz$ - so here I want to multiply the $\phi$ component by $r$,... but if I do that then $E$ comes out incorrectly.

So my question is - have I made a mistake in the calculation above? Or if it is correct then is the definition of the magnitude of the electric field equal to the equation below...

$$|E|=\sqrt{\left({\partial V \over \partial r}\right)^2 + \left({1\over r}{\partial V \over \partial \phi}\right)^2+ \left({\partial V \over \partial z}\right)^2}$$

Maybe another way to ask this question is to ask if $\hat{\phi}$ is a vector with unit length of angle - or unit length of distance.

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  • $\begingroup$ @Broly - thanks for help trying to edit my messed up equation at the end - problem was that I got the maths of it wrong and then updated it myself so that when your edit came in it actually had the old incorrect equation - so sorry for rejecting your edit $\endgroup$ – tom Apr 13 at 21:03
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    $\begingroup$ @DanielTuzes - many thanks for the helpful edit. Sorry I was not online to approve the edit, but it got approved anyway. $\endgroup$ – tom Apr 14 at 14:57
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The dimension of $V$ is $V$ as volts. The dimension of its gradient is $\left[ {\nabla V} \right] = V/m$, so in the notation $$ - \nabla V = - \left( {\frac{{\partial V}}{{\partial r}}{\hat{\mathbf{\hat r}}} + \frac{1}{r}\frac{{\partial V}}{{\partial \phi }}{\mathbf{\hat \phi }} + \frac{{\partial V}}{{\partial z}}{\mathbf{\hat z}}} \right)$$ the dimension of ${{\mathbf{\hat r}}}$, $\hat {\mathbf{\phi }}$ and ${{\mathbf{\hat z}}}$ are all 1. ${{\mathbf{\hat \varphi }}}$ points perpendicular to the actual value of ${{\mathbf{\hat r}}}$.

You did the calculation correctly.

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  • $\begingroup$ thanks for your helpful answer, I appreciate it - I thought I was going mad... $\endgroup$ – tom Apr 13 at 21:44
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    $\begingroup$ It is helpful to see the thing about the dimensions of the gradient function. I appreciate it. $\endgroup$ – tom Apr 13 at 21:45
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Ok from this page it looks like $\hat{\phi}$ is indeed a unit vector of length - which in cartesian coordinates would be represented by

$$ \hat{\phi} = -\sin(\phi) \hat(i) + \cos(\phi) \hat{j}$$

which means that the equation for the electric field in the question above is correct...

... and it also means that there is inconsistency between $d\phi$ and $\hat{\phi}$ because the first is a infinitessimal bit of angle and needs to be multiplied by $r$ to get a length, whereas the second is unit length and does not need to be multiplied by $r$ to make a length!

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