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I am wondering how the author rationalizes the removal of absolute value bars around the quotient argument of a natural logarithm. My take on this is that the potential at point $b$ MUST be greater than the potential at point $a$, or else the removal of the absolute value bars obtained through anti-differentiation would be unwarranted. This question has actually come up several times in problems relating to potential difference and any clarification would be much appreciated. Here are links to the example from my text.



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up vote 4 down vote accepted

He is taking the log of the quotient of two radii. Since radii are always non-negative, the quotient is non-negative, so the absolute value can be omitted.

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Ahh, I see. Thank you. Is there any particular reason he integrated from point a to b? Could you integrate from b to a? – ao2130 Mar 16 '14 at 19:43
@ao2130: You are looking for the potential difference from $a$ to $b$, so you should integrate from $a$ to $b$. You could go $b$ to $a$, but you'd have to take a minus sign into account: $\int_a^b=-\int_b^a$. – Kyle Kanos Mar 17 '14 at 1:36

There is no particular reason for integrating from a to b, or b to a. The important thing is the difference in potential between two points. However when integrating be careful with changing the signs (add a negative sign for the integral of b to a). So $V_{a} - V_{b} = V_{b} - V_{a} = \Delta V$

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Thank you very much. – ao2130 Mar 16 '14 at 19:53

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