# Calculating electric field in cylindrical symmetry- - Trying to understand electric field/potential in 3D with cylindrical coordinates

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

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

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.

• @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 – tom Apr 13 at 21:03
• @DanielTuzes - many thanks for the helpful edit. Sorry I was not online to approve the edit, but it got approved anyway. – tom Apr 14 at 14:57

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.

Ok from this page it looks like $$\hat{\phi}$$ is indeed a unit vector of length - which in cartesian coordinates would be represented by
$$\hat{\phi} = -\sin(\phi) \hat(i) + \cos(\phi) \hat{j}$$
... 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!