This is a trivial question probably, i have a doubt when considering the signs for the Lagrangian for a charged particle in an electromagnetic field.

Considering that the Lagrangian for a charged particle in an electromagnetic field is given as
$$L = \frac{m v^2}{2} + q \vec{v} \cdot \vec{A} - q \phi \ \ \ \  \text{or} \ \ \ \ L = \frac{m v^2}{2} - q \vec{v} \cdot \vec{A} + q \phi$$
And that the equation of motion for this Lagrangian is associated with the Lorentz force as
$$m \dot{v} = q(E + v \times B) \ \ \ \ or \ \ \ \ m \dot{v} = -q(E + v \times B)$$
My doubt is:

If I have the Lagrangian as
$$L = \frac{m v^2}{2} - q \vec{v} \cdot \vec{A} - q \phi$$
that gives the equation of motion as
$$m \dot{v} = q(-E + v \times B)$$

Does this equation of motion still represents the Lorentz force even with the difference in signs, or $E$ and $B$ must have the same sign?

Is it correct to interpret this Lagrangian as a charged particle in an electromagnetic field?