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.

 

[palaphys]

Your two methods look correct.

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

I'm not able to edit it right now, but thanks for pointing it out.

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

how is that possible? even their solution has included the effect of gravity right

 

[TSny]

how is that possible? even their solution has included the effect of gravity right

Yes, they correctly accounted for gravity. But, they did not account for the change in chemical energy $U_{chem}$ of the battery. Their solution predicts an imaginary value for the final speed of the plate for the case where $g = 0.$

This is a good problem for demonstrating the importance of clearly defining your "system".

If we take the capacitor, the earth, and the battery as the system, then we have an isolated system. So, $\Delta E_{sys} = 0$. Then $$\Delta U_{cap} + \Delta U_{grav} + \Delta U_{chem} + \Delta K = 0.$$
If we consider the capacitor and the earth as the system, then the battery acts as an external agent doing work on the system. Then, $$W_{bat} = \Delta U_{cap} + \Delta U_{grav} + \Delta K.$$
Other possibilities are:
(a) Choose the capacitor to be the system,
(b) Choose just the top plate of the capacitor to be the system.

What would the work-energy equation look like for these two choices?

If set up correctly, all of these different choices for the system will yield the same answer for the speed of the plate.