Speed and direction from one energy line

An isolated charge moving through a static field has only the electrostatic interaction acting on it, so mechanical energy is conserved:

$$\Delta K + \Delta U = 0 \qquad \Longrightarrow \qquad \Delta K = -\Delta U = -q\,\Delta V = q(V_{\text{start}} - V_{\text{end}}).$$

Everything on the right is read off a potential map. No force is computed, no acceleration is named, and the trajectory is never needed: only the two endpoint potentials and the charge.

The standard procedure is four steps and it does not change from problem to problem.

1. Find $V$ at the start and $V$ at the end, from $kq/r$, from $V = Ed$, or from a graph.
2. Subtract in the order end minus start to get $\Delta V$.
3. Multiply by the charge with its sign to get $\Delta U = q\Delta V$.
4. Set $\Delta K = -\Delta U$ and solve for the speed with $K = \tfrac12 mv^2$.

At atomic scale the electron volt makes step 4 nearly free: an elementary charge crossing $1$ V picks up $1$ eV $= 1.6\times10^{-19}$ J, so an electron accelerated through $1000$ V arrives with $1000$ eV. Convert to joules only when a speed is wanted.

The word "downhill" is a statement about energy, and $V$ is not an energy. Since $U = qV$, the hill a charge feels depends on the sign of that charge:

1. Positive charge. $U$ tracks $V$, so the charge accelerates toward LOWER potential. Downhill in $V$ is downhill in $U$.
2. Negative charge. $U = qV$ flips the graph upside down, so $U$ is lowest where $V$ is HIGHEST. The charge accelerates toward higher potential.

Both statements are the same statement: every charge accelerates toward lower $U$. Only the translation into $V$ changes.

So an electron released between plates at $0$ V and $200$ V moves toward the $200$ V plate and gains kinetic energy doing it. Sending it the other way, toward $0$ V, is uphill and costs energy, which is why a claim that it "falls to the low plate and speeds up" is a claim that energy appeared from nowhere.

The same flip is visible in the field picture and the two must agree. $\vec{E}$ points from high $V$ toward low $V$, always. The force is $\vec{F} = q\vec{E}$, so on a negative charge the force is opposite the field, which is up the potential gradient. If your energy answer and your force answer disagree about direction, one of them dropped a sign, and it is usually the one that never wrote the charge's sign down.

$v^2 = v_0^2 + 2as$ is derived on the assumption that $a$ is constant. Near a point charge the field goes as $1/r^2$, so the force and the acceleration change at every instant, and the formula does not apply. Computing $a = qE/m$ at the starting point and running it across the whole trip is not an estimate; it is an answer to a different problem, and it typically lands a factor of three off.

Energy does not care. $\Delta K = -q\Delta V$ is exact whatever the field does in between, because the potential difference already contains the integral of the varying force:

$$q\,\Delta V = -q\int \vec{E}\cdot d\vec{l}.$$

Take a proton starting at rest $2.0$ cm from a fixed $+10$ nC charge and moving out to $6.0$ cm. The potentials are $kQ/r = 4495$ V and $1498$ V, so the proton arrives with $K = e(2997\ \text{V}) = 4.8\times10^{-16}$ J. Two lookups and a subtraction, with no trajectory and no integral in sight.

Kinematics is available again the moment the field is uniform: between parallel plates, well inside a long capacitor, $a = qE/m$ is genuinely constant and the constant-acceleration equations are exact. The test to run before reaching for them is not "does the charge start from rest" or "is the path straight", it is "is the field the same at every point on the path".

In electrostatic equilibrium the field inside a conductor is zero, because any interior field would drive the free charges until it was cancelled. Feed $E = 0$ into $E_x = -dV/dx$ and what comes out is

$$\frac{dV}{dx} = 0 \qquad \Longrightarrow \qquad V = \text{constant},$$

not $V = 0$. The zero of a derivative says the function is flat; it says nothing about the function's value. Those are different claims, and conflating them would make every isolated charged sphere a ground.

For an isolated conducting sphere of radius $R$ carrying $Q$, the potential is

$$V(r) = \frac{kQ}{r} \ \ (r \ge R), \qquad V(r) = \frac{kQ}{R} \ \ (r \le R).$$

Outside, it falls as $1/r$ exactly as if all the charge were at the centre. At the surface it reaches $kQ/R$, and from there inward it runs level all the way to the centre. A $2.0$ nC sphere of radius $10$ cm sits at $180$ V from its surface to its core.

The same reasoning covers a hollow shell. Inside the cavity of a charged conductor with no enclosed charge, $E = 0$, so $V$ is constant there too and equal to the shell's own potential. A charge released inside such a cavity stays put: no field, no force, and the potential it sits at is large, not zero.