Why electic potential on the center of a disk axis = 0

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SUMMARY

The electric potential (E.P) at the center of a charged disk is zero due to the integration of the electric potential formula, specifically dv = 2k(q/a^2)((r^2)+(x^2))^(.5)dr. In this equation, 'a' represents the radius of the disk, 'r' is the radius of the elemental charge, and 'x' is the distance from the disk to the point of interest. The integration limits for 'v' change from 0 to v, and for 'r' from 0 to 'a', confirming that when r equals zero, v equals zero. The concept of electric potential is relative, allowing for the zero potential to be defined at any logical point, such as the center of the disk.

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moamen811
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when i saw how to get the electic potential of a charged disk at a point on its axis i found that the E.P on center of the disk =0 as
the integration of the eq:
dv = 2k(q/a^2)((r^2)+(x^2))^(.5)dr...(a) is the radius of the disk , (r) is the radius of the element and (x) is the distance between the disk and the point we get the E.P at

i found that the integration of v changed from (0---> v) and r changed from (0--->a)
and that means that when r=zero v=zero ...why??


sorry if my english is not good my native lang. is not english ...
 
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I am finding it hard to read your post, but I will say one thing:

A potential (be it electric or something else) is only meaningless if you look at a potential difference. For that reason, you can take your zero potential anywhere you want. You can take it in the center of the disc, or on the moon; the final answer should be the same.
However, calculations or often much simpler if you take a logic place, such as the center of the disc, or at infinity.
 

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