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Here in the first image it is said $\displaystyle \int_a ^b E\cdot \mathrm dl\,$ is $0 $:

enter image description here

Here in the second image it looks like $\displaystyle \int_a ^b E\cdot \mathrm dl\,$ is not $0$:

enter image description here

If field is conservative which it is here, then closed line integral is 0 which is the case in the first image, but the potential is also a line integral so why is it not 0

Please help me understand this!

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The integral around any closed path is zero. But $\vec{a}$ and $\vec{b}$ are not the same point: the path is not closed, so the integral does not have to be zero.

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    $\begingroup$ But a and b are different points in the first image, they say (DJ Griffiths) 'a' is $R_a$ from the charge at origin and b is $R_b$ from the origin, then how can a and b be equal ? So how can the line integral be 0 there ? $\endgroup$ – theenigma017 Aug 19 at 9:49
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    $\begingroup$ Yes, and then he gets an expression for that integral in general in terms of $r_a$ and $r_b$ and shows that it is zero if $\vec{a}$ & $\vec{b}$ are the same point: it's not always zero. $\endgroup$ – tfb Aug 19 at 10:05
  • $\begingroup$ It will be 0 if a and b are at the same point for any field wouldn't it ? because it's just subtracting same values to get 0 ? $\endgroup$ – theenigma017 Aug 19 at 11:19
  • $\begingroup$ imgur.com/a/ruc0gV4 In the figure a and b can be different points and $R_a$ and $R_b$ can be equal too ? Where is the closed path here ? All i see is a curvy line from a to b. $\endgroup$ – theenigma017 Aug 19 at 11:22
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    $\begingroup$ Yes, they can be different points: different points can have the same potential. However identical points must have the same potential. This is part of the proof that there is in fact a well-defined potential. $\endgroup$ – tfb Aug 19 at 11:25

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