Electric Field under Time reversal

The electric force acting on a particle can be expressed as:

$$vec{F} = q vec{E} $$

And also as:

$$vec{F} = m frac{d^2x}{dt^2}$$

The charge is a scalar, so $vec{F}$ behaves as $vec{E}$ because of the first equation and it doesn't change sign under time reversal due to the second one. So $vec{E}$ doesn't change sign under time reversal.

Let's consider the simplest case, a one dimensional problem where the electric field is constant and equal to $E_0$. Then your equations are:

$$qE_0 = frac{d^2x}{dt^2}$$

And you get:

$$v equiv frac{dx}{dt} = qE_0 t + A$$

Where $A$ is a constant. So, you can see from this equation that under time reversal the velocity changes sign which is what you are interested in

Under the time reversal the velocity changes the direction, while the elctric field and therefore force, doesn't. That means that if in normal time a particle is moving in one direction and accelerating (force in the same direction as the velocity), then after time reversal we'll see a particle moving inan opposite direction and slowing down (force opposite to the velocity), as it should be.

If we consider a force orthogonal to the velocity, causing a circular motion it will also work. Under the time reversal the particle will be seen circling in the opposite direction, but the center of the circle will still be at the same place. That means that the force, which points at the center of the circle, stays the same, nit doesn't change the direction.