Express E due to a point charge at the origin

[leeban7]

Homework Statement

Express the electric field E due to a point charge q at the origin in cylindrical polar coordinates.

Homework Equations

The Attempt at a Solution

Know that E = q / 4*pi*epsilon_0*r^2 in the r-direction, which is the answer in spherical coordinates. How we we swap to cylindrical?

 

[BvU]

Hello leeban,

Same as when we would have to change to Cartesian coordinates:
We simply express the $\vec E$ in terms of $E_\rho$, $E_\phi$ and $E_z$

 

[leeban7]

Do you recommend swapping to Cartesian and then to cylindrical? I know the linear transforms for those, but not from spherical to cylindrical

 

[BvU]

I do not recommend that. For one, you already have that $\phi$ in the one is equal to $\phi$ in the other system (*)
And for the $z$, yes, you have to do that, because Cartesian $z$ = cylindrical $z$.
And from $r$ to $\rho$ is a breeze.

(*) you want to beware that those stupid mathematicians chose to confuse everyone by swapping $\theta$ and $\phi$

[edit] proof of this claim: check out how wikipedia makes a mess of your exercise.

 

[BvU]

which is the answer in spherical coordinates

Not the full answer, though ! You only mention $E_r$

 

[leeban7]

Due to symmetry isn't the electric field on dependent on r?

 

[BvU]

The magnitude is, yes. But $\vec E$ is a vector: it also has a direction (along $\hat r$) and you only give the radial component without stating the other two.

 

[leeban7]

Would I be correct in then saying that
\vec E = q/4*pi*epsilon_0*r^2 \hat r + 0 \hat omega + 0 \hat phi

 

[BvU]

Would I be correct in then saying that
$$\vec E = {q\over 4\pi\varepsilon_0 r^2} \, \hat r + 0 \, \hat \phi + 0 \, \hat \theta$$

Yes, that (somewhat edited ^* ) is the full expression I hinted at. Hopefully makes it easier for you to convert to cylindrical.

^* $$\vec E = {q\over 4\pi\varepsilon_0 r^2} \, \hat r + 0 \, \hat \phi + 0 \, \hat \theta$$