Falling capacitor connected to constant voltage

[palaphys]

Homework Statement

An air filled parallel plate capacitor with plate separation $d$ and area $A$ is connected to a constant voltage source $V$. The plates are horizontal and the top plate is free to move vertically without rotation. The system is in vacuum and friction is absent. If the upper plate is released from a height $d$, what is its speed when the separation becomes $d/2$?

Relevant Equations

Kinematics+Capacitor equations

I have tried to proceed with two methods, which agree with each other
i) Energy analysis
If we use work energy theorem,
$\sum W = \Delta K$

So,
$W_{\text{bat}} + W_{\text{field}} + W_{\text{grav}} = K_f \quad \text{(at } x = d/2 \text{)}$

Battery work:
$W_{\text{bat}} = V \cdot (C_i V) = C_i V^2 = \frac{\varepsilon_0 A V^2}{d}$

Change in electrostatic energy:
$\Delta U_{\text{elec}} = U_f - U_i = \frac{1}{2} C_f V^2 - \frac{1}{2} C_i V^2 = \frac{1}{2}(2C_i)V^2 - \frac{1}{2} C_i V^2$
$\Delta U_{\text{elec}} = C_i V^2 - \frac{1}{2} C_i V^2 = \frac{1}{2} C_i V^2$

Change in gravitational potential energy:
$\Delta U_{\text{grav}} = U_{g,f} - U_{g,i} = Mg\left(\frac{d}{2}\right) - Mgd = -Mg\frac{d}{2}$

Substituting in the work–energy theorem:
$M v^2 = C_i V^2 + Mg d$
$v^2 = \frac{C_i V^2}{M} + g d$

Final expression using $(C_i = \frac{\varepsilon_0 A}{d}\)$
$\boxed{v = \sqrt{\frac{\varepsilon_0 A V^2}{M d} + g d}}$

ii) force method:
$\text{Let } x \text{ be the downward displacement (so } x=0\text{ at start, } x=d/2 \text{ at end).}$

Net downward force:
$F_{\text{net}} = mg + \frac{1}{2}\frac{\varepsilon_0 A V^2}{(d - x)^2}$
then I applied 2nd law:
$m\,v\frac{dv}{dx} = mg + \frac{1}{2}\frac{\varepsilon_0 A V^2}{(d - x)^2}$

integrating from initial $\((x=0,\,v=0)\)$ to final $\((x=d/2,\,v=v)\)$:
$\int_{0}^{v} m v\,dv = \int_{0}^{d/2} \left( mg + \frac{1}{2}\frac{\varepsilon_0 A V^2}{(d - x)^2} \right) dx$

LHS
$\frac{1}{2} m v^2$

RHS
$mg\frac{d}{2} \;+\; \frac{1}{2}\varepsilon_0 A V^2 \int_{0}^{d/2} \frac{dx}{(d - x)^2}$

integration:
$\int_{0}^{d/2} \frac{dx}{(d - x)^2} = \left[-\frac{1}{d-x}\right]_{0}^{d/2} = -\frac{1}{d - d/2} + \frac{1}{d} = \frac{1}{d}$

Thus:
$\frac{1}{2} m v^2 = mg\frac{d}{2} + \frac{1}{2}\frac{\varepsilon_0 A V^2}{d}$

So
$v^2 = \frac{\varepsilon_0 A V^2}{m d} + g d$
$\boxed{v = \sqrt{\frac{\varepsilon_0 A V^2}{m d} + g d}}$

I hope both my methods are right. however, in the solution presented right below the problem, work done by the battery is not accounted for at all. Who is wrong here and why?(see below)

 

[TSny]

Your two methods look correct.

I did catch a spot where you have two compensating errors:

Check the signs in this calculation. However, your result of $\frac 1 d$ is correct.

To see that their answer can't be correct, suppose the experiment is done in zero gravity.