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## Main Question or Discussion Point

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 ....

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 ....