Build the hill out of qV, not V

Objects move toward lower potential energy, and potential energy is $U = qV$. For a positive charge that agrees with the language of downhill: it speeds up as it moves toward lower potential. For a negative charge the factor of $q$ flips the landscape over.

Release an electron beside a $100$ V plate with a $0$ V plate across the gap:

$$U = qV = (-e)(100\ \text{V}) < 0 \quad \text{at the high-potential plate},$$

which is the lowest $U$ available to it. So the electron does not race across to the $0$ V plate; it is already at the bottom of its own hill, and moving toward $0$ V would cost it energy. An electron accelerates toward higher potential.

The working rule: find the direction in which $U$ decreases, not the direction in which $V$ decreases, and the motion follows. Those are the same direction only when $q > 0$.

The one line that answers most of these problems:

$$\Delta K = q\left(V_{\text{start}} - V_{\text{end}}\right), \qquad \text{so} \qquad \tfrac{1}{2}mv^2 = q\,\Delta V \ \text{from rest}.$$

Nothing in that derivation assumed anything about the path or about how the field varies along it. Get the two potentials however you can, from $kq/r$ near a point charge or from the plate values in a capacitor, subtract, and multiply by the charge with its sign.

A useful unit falls out. A charge $e$ moved through $1$ V gains $1$ electronvolt, $1.6\times10^{-19}$ J, which is why particle energies are quoted in eV and keV.

Compute $a = qE/m$ at the release point beside a point charge and the number is correct, for that instant. Move the charge $3$ cm and $E$ has changed, because $E = kq/r^2$ depends on where you are. So $v^2 = v_0^2 + 2ad$ has no constant $a$ to use and the answer it produces is wrong.

1. Uniform field, between parallel plates. The force really is constant, so $a = qE/m$ holds for the whole trip and all the constant-acceleration relations apply. Launch a charge sideways and it traces a parabola, exactly as a projectile does.
2. Anywhere near a point charge or any non-uniform field. Use energy. Get $V$ at the start and at the end, subtract, multiply by $q$.

The tell for the misapplication is a solution that computes one acceleration and then uses it over a distance across which the problem has clearly told you the separation changed.

A charge $q$ is released from rest at distance $r_1$ from a fixed source $Q$, and you want its speed at $r_2$. The whole solution is four steps and no kinematics.

1. $V_1 = kQ/r_1$ and $V_2 = kQ/r_2$, each with the sign of $Q$.
2. $\Delta K = q(V_1 - V_2)$, with the sign of $q$.
3. Started from rest, so $\tfrac12 mv^2 = \Delta K$.
4. $v = \sqrt{2\,\Delta K/m}$.

Two checks worth running. If the pair is like-signed and moving apart, $\Delta K$ must come out positive, because they repel. If $\Delta K$ comes out negative for a charge you released from rest, the charge cannot get there at all, and that is a real answer rather than an error: it means the charge turns around first.