What is $\mathcal{O}$ in integration of potential?
I saw following equation in Griffiths EM $$V(r)=-\int_\mathcal{O}^r \vec E \cdot d\vec l$$
While the surface was closed but not symmetrical circular. At first I was thinking $\mathcal O$ was representing closed surface and $r$ was representing radius of that circle. But when I noticed the circle isn't symmetrical I claimed that my thought was wrong. The circle was started at point A and end at point B, After looking at that non-symmetrical circle I thought Griffiths will write an integrand like this : $$V(r)=-\int_A^B \vec E \cdot d\vec l$$ but his method confused me.
1 answer
Actually $\mathcal O$ is some standard point and $r$ is where potential depends on.
$$V(b)-V(a)=-\int_{\mathcal{O}}^B \vec E \cdot d\vec l +\int_{\mathcal{O}}^A \vec E \cdot d\vec l $$ $$ =-\int_{\mathcal{O}}^B \vec E \cdot d\vec l-\int^{\mathcal{O}}_B \vec E \cdot d\vec l$$ $$=-\int_A^B \vec E \cdot d \vec l$$
which was completely related to OP's equation. But the problem was the OP was talking about a potential point while it's valid for potential differences.
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