# Resolution in cylindrical co-ordinates

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

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

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.

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

Edit: See the figure.

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