The circulation of the electric field gives the potential difference, but is it :
$$V_B-V_A = \int_A^B\vec{E}.\vec{dOM} \hspace{1.5cm} (1)$$
or
$$V_B-V_A = - \int_A^B\vec{E}.\vec{dOM} \hspace{1cm} (2)$$
I would say $(2)$ but I'm not sure because usually in math, $ \int_a^b f(x)dx = f(b)-f(a)$