The circulation of the electric field gives the potential difference, but is it :

$$V_B-V_A = \int_A^B\vec{E}.\vec{dOM} \hspace{1.5cm} (1)$$


$$V_B-V_A = - \int_A^B\vec{E}.\vec{dOM} \hspace{1cm} (2)$$

I would say $(2)$ but I'm not sure because usually in math, $ \int_a^b f(x)dx = f(b)-f(a)$

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    $\begingroup$ (2) is correct because the field is minus the gradient of the potential. $\endgroup$ – Robin Ekman Mar 12 '15 at 8:43

It depends on how you define the potential. In mechanics one has the convention $$ F = -\nabla U$$ Since the electric field exerts a force via $F=qE$ it is only natural to apply this convention in electrostatics too. In this way the electric potential $V$ can be directly interpreted as mechanical potential energy $U=qV$.

Option number (2) is therefor the common choice.


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