I'm trying to resolve this problem in cylindrical co-ordinates.

I have two charges placed along the $z$-axis separated by a distance $a$.

enter image description here

Why is that the resolution is made only in the direction of $\rho$ (in the figure $r$) and $z$ direction and not in the direction of $\phi$.

I was able to resolve the problem in the cartesian co-ordinates and from there obtain the same results.

$$\vec E_1 =\frac{q}{4\pi \epsilon_o}\frac{x \hat i_x+y \hat i_y-\frac{a}{2} \hat i_z}{(x^2+y^2+\frac{a^2}{4})^{\frac{3}{2}}}$$

$$\vec E_2 =\frac{q}{4\pi \epsilon_o}\frac{x \hat i_x+y \hat i_y+\frac{a}{2} \hat i_z}{(x^2+y^2+\frac{a^2}{4})^{\frac{3}{2}}}$$

and when I add I get

$$\vec E_1 + \vec E_2 =\frac{q}{4\pi \epsilon_o}\frac{1}{(x^2+y^2+\frac{a^2}{4})^{2}}(\frac{x}{\sqrt{x^2+y^2}} \hat i_x + \frac{y}{\sqrt{x^2+y^2}} \hat i_y) \\=\frac{q}{4\pi \epsilon_o}\frac{1}{(x^2+y^2+\frac{a^2}{4})^{2}}(cos\phi \hat i_x + \ sin\phi \hat i_y) \\=\frac{q}{4\pi \epsilon_o}\frac{1}{({\rho}^2+\frac{a^2}{4})^{2}}\hat i_{\rho}$$

where $\phi$ is the angle made with the $x$-axis and since $\rho^2=x^2+y^2$ and $\hat i_{\rho}=cos\phi \hat i_x +sin\phi \hat i_y $

But I want to understand why $\phi$ component was not included in the resolution with cylindrical co-ordinates as is shown in the figure.


1 Answer 1


The field is radially outward, so there can't be a $\hat{\phi}$ component.

Edit: See the figure.

enter image description here

  • $\begingroup$ I do know that....but how is there no $\hat \phi$ component? $\endgroup$
    – Orpheus
    Commented Jun 2, 2021 at 6:10
  • $\begingroup$ I mean does it cancel out somewhere? $\endgroup$
    – Orpheus
    Commented Jun 2, 2021 at 6:11
  • $\begingroup$ See the figure, The vertical component get canceled. $\endgroup$ Commented Jun 2, 2021 at 7:06

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