Conservation of Electric Energy

With only the electrostatic interaction acting, $\Delta K + \Delta U = 0$, so $\Delta K = -q\Delta V$. That single line gives the speed at any point from the two endpoint potentials, with no force, acceleration or trajectory required. Because $U = qV$, a positive charge accelerates toward lower potential and a negative charge accelerates toward HIGHER potential: both move toward lower $U$, and only the translation into $V$ flips. Inside a conductor in equilibrium $E = 0$, which by $E_x = -dV/dx$ makes $V$ constant at the surface value $kQ/R$, not zero.

Three errors dominate. Sending every charge from high $V$ to low $V$, which is right for a proton and backward for an electron, and produces kinetic energy that the ledger cannot pay for. Reaching for $v^2 = v_0^2 + 2as$ near a point charge, where the acceleration changes at every position, when $q\Delta V$ would have answered it exactly in one line. And reading $E = 0$ inside a conductor as $V = 0$, when the vanishing of a derivative makes the function flat rather than zero, so the interior of a charged sphere is pinned at its surface potential all the way to the centre.