# If electric field is conservative then line integral along closed path is zero, then why is potential not zero?

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

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

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

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.
• 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 ? – theenigma017 Aug 19 at 9:49
• 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. – tfb Aug 19 at 10:05
• 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. – theenigma017 Aug 19 at 11:22