Showing E.dl is 0 - Why cylindrical coordinates?

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emhelp100
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Homework Statement


A point charge +Q exists at the origin. Find [itex]\oint[/itex] [itex]\vec{E}[/itex] [itex]\cdot \vec{dl}[/itex] around a circle of radius a centered around the origin.

Homework Equations

The Attempt at a Solution


The solution provided is:
[itex]\vec{E} = \hat{\rho}\frac{Q}{4\pi E_0a^2}[/itex]
[itex]\vec{dl}=\hat{\phi}\rho d\phi[/itex]
Why are cylindrical coordinates being used here?
 
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jtbell said:
In principle, one can use any coordinate system for this problem. However, it's simpler to use spherical coordinates in this case. To see why, write ##\vec E## and ##\vec {dl}## in rectangular (Cartesian) coordinates.
Not really sure what it would be in rectangular coordinates...
 
Start by drawing a diagram that shows ##\vec E## and ##\vec {dl}## at a point a distance a from the origin and at some arbitrary angle φ with respect to the x-axis. Then resolve both of those vectors into x- and y- components.
 
emhelp100 said:

Homework Statement


A point charge +Q exists at the origin. Find [itex]\oint[/itex] [itex]\vec{E}[/itex] [itex]\cdot \vec{dl}[/itex] around a circle of radius a centered around the origin.

[snip]

Why are cylindrical coordinates being used here?

As @jtbell states, any coordinate system can be used.
You'll get the same result... but one choice of coordinates might be easier than another.
Generally speaking, if you choose a coordinate system that exploits a symmetry in the problem, then your math problem (your integration problem) will be simpler.
The point charge at the origin describes something with spherical symmetry.
The circle centered around the origin has cylindrical symmetry.

(Would you rather calculate the circumference of a circle in rectangular coordinates or in polar coordinates?)