Limits of integration for electric potential

Calpalned · Oct 22, 2015

In the example in the picture, we can see that they chose the limits of integration to be from 0 to R_0. Why didn't they choose x (that is, from 0 to x)? Isn't that what we normally integrate over when we find potential energy and electric fields?
Thank you

Textbook page, Example 23-9, showing the derivation of the potential at an axial point due to a uniformly charged disk, with the integral of dq over the disk and the result V = Q/(2 pi epsilon0 R0) times (sqrt(x^2 + R0^2) minus x), plus a note for x much greater than R0, and Figure 23-15 showing a disk with a point P on its axis.

mukundpa · Oct 22, 2015

Here we are adding potential at P due to different ring elements. The disk is considered as large numbers of thin rings with radii varying from 0 to R₀ hence the potential due to a ring element is integrated for radius 0 to R₀.

Calpalned · Oct 22, 2015

That makes sense... so integrating from 0 to x would be invalid?

mukundpa · Oct 22, 2015

Yes, actually integration is the sum and we are here adding nothing which is varying with x. The distance x of P is a constant.

Calpalned · Oct 22, 2015

mukundpa said:

Yes, actually integration is the sum and we are here adding nothing which is varying with x. The distance x of P is a constant.

Ok thank you. I understand now

mukundpa · Oct 22, 2015

welcome.

Limits of integration for electric potential

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