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I was answering this question and even tho i know intuitively that the radial component of the eletric field will cancel out by simmetry, i could not get that in the integral calculation. I´m asking if you can help me figure out what mistake i did in the exercise. Thank you!

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  • $\begingroup$ If P is a point on the z axis, why is it indicated as a function of φ? $\endgroup$
    – R.W. Bird
    Commented Nov 4, 2020 at 14:08
  • $\begingroup$ I think because P has a radius of 0, φ can have any possible value, so if we assume this we can cancel out the component in the direction of φ. $\endgroup$
    – João Marques
    Commented Nov 4, 2020 at 14:12

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Assuming that the charge is a disk in the xy plane and centered at the origin, then:

d$E_­z$ = z dq/[4π$ε_o(z^2 + r^2)^{1.5}$] and taking dq as a ring of charge in the xy plane, then:

dq = ((20E-6)/r)(2πr)dr.

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  • $\begingroup$ Thanks for the reply, but what i am trying to figure out is when i go by the more fundamental approach, the radial component is not cancelling in the definite integral, even tho that is what happens. $\endgroup$
    – João Marques
    Commented Nov 4, 2020 at 14:47
  • $\begingroup$ It is my understanding that an integral is a scaler sum. You cannot integrate vectors; only components of vectors. The 6 m is a z component, so that term can be integrated, but the unit radial vector must be broken into x and y components and each of these becomes a double integral on r and φ. Integrating a sine or cosine from zero to 2π gives a zero result. $\endgroup$
    – R.W. Bird
    Commented Nov 4, 2020 at 18:53
  • $\begingroup$ Yes, i understand now thank you very much! $\endgroup$
    – João Marques
    Commented Nov 4, 2020 at 20:30

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