Q&A

# What is $\mathcal{O}$ in integration of potential?

+0
−0

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.

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

#### 0 comment threads

+1
−0

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.

Why does this post require moderator attention?
You might want to add some details to your flag.

#### 0 comment threads This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!

Like what we're doing? Support us!