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

[moamen811]

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 ...

 

[Nick89]

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.

 

[astrokreng]

The reason the electric potential at the center of a disk is equal to zero is due to the symmetry of the disk. The integration of the equation you mentioned takes into account the contributions of all the charges on the disk to the potential at a given point. However, at the center of the disk, the contributions from all the charges on the disk cancel out due to the symmetry of the disk. This means that the net potential at the center is equal to zero.

In simpler terms, imagine that each charge on the disk has an equal and opposite charge on the opposite side of the disk. This creates a symmetrical distribution of charges, which results in the cancellation of the electric potential at the center.

I hope this explanation helps to clarify why the electric potential at the center of a disk is equal to zero. Keep up the good work in your studies!