Answer
$mgd/q$
Work Step by Step
We have two cases here:
First, the bead is fired under the free-fall acceleration when the two-plate capacitor is uncharged. See the left figure below.
The energy is conserved, so
$$K_i+U_{ig}+U_{ie}=K_f+U_{fg}+U_{fe}$$
there are no electric potentials yet, so
$$K_i+U_{ig}+0 =K_f+U_{fg}+0$$
at the highest point the bead stops before falling back, so $K_f=0$
$$K_i+U_{ig} = 0+U_{fg} $$
Let the bottom plate as $y=0$, so $U_{ig}=0$
$$K_i+0= 0+U_{fg} $$
$$\frac{1}{2} \color{red}{\bf\not} mv_0^2= \color{red}{\bf\not} m g h $$
Hence,
$$h=\dfrac{v_0^2}{2g}\tag 1$$
Second, the bead is fired under the free-fall acceleration plus the electric force acceleration due to the electric field between the two plates when the two-plate capacitor is charged.
The energy is conserved, so
$$K_i+U_{ig}+U_{ie}=K_f+U_{fg}+U_{fe}$$
Let the bottom plate as $y=0$, so $U_{ig}=0$
$$K_i+0+U_{ie}=0+U_{fg}+U_{fe}$$
$$\frac{1}{2} mv_0^2 =\dfrac{mgh}{2}+\Delta U_{e} $$
$$\frac{1}{2} mv_0^2 =\dfrac{mgh}{2}+q(V_f-V_i) $$
$$\frac{1}{2} mv_0^2 =\dfrac{mgh}{2}+q(Es-0) $$
where $s=h/s$
$$\frac{1}{2} mv_0^2 =\dfrac{mgh}{2}+\dfrac{qEh}{2} $$
Recall that the electric field inside a two-plate capacitor is constant and uniform and is given by $E=\Delta V_C/d$
$$\frac{1}{ \color{red}{\bf\not} 2} mv_0^2 =\dfrac{mgh}{ \color{red}{\bf\not} 2}+\dfrac{qh\Delta V_C}{ \color{red}{\bf\not} 2d} $$
Hence,
$$ mv_0^2 -{mgh} =\dfrac{qh\Delta V_C}{ d} $$
$$ \Delta V_C= \left[mv_0^2 -mgh\right]\dfrac{d}{qh} $$
$$ \Delta V_C= \dfrac{mv_0^2d}{qh} -\dfrac{mg d}{q } $$
Plugging from (1),
$$ \Delta V_C= \dfrac{mv_0^2d}{q } \dfrac{2g}{v_0^2} -\dfrac{mg d}{q } $$
$$ \Delta V_C= \dfrac{2m g d}{q } -\dfrac{mg d}{q } $$
$$\boxed{ \Delta V_C= \dfrac{mg d}{q } }$$
