Rotary variable capacitor

Given the limited solution choices, we can simply thing about the work that it would take to make the rotation. Simply put, the positive charge on one plate attract the negative charge on the other plate. Rotating by 90 degrees pulls the charges apart and would take positive work, so the system energy must increase (i.e. not stay the same or halve).

(You may be worried that the charge density will move around during the rotation, and it no doubt will due to the changing geometry of the capacitor. As I'm writing this, I can't think or a really qualitative reason why we can discount this. Reply if you have a simple qualitative argument to this effect.)

Relevant wiki: Capacitors

The electrical charge $Q$ equally distributed over the overlapping area $A(\varphi) = (\pi - \varphi) R^2$ of both disks. Gauss' law results $\begin{aligned} & & \oint \vec E \cdot d\vec A &= E A(\varphi) = \frac{Q}{\varepsilon_0}\\ \Rightarrow & & E &= \frac{Q}{\varepsilon_0 (\pi - \varphi) R^2} \\ \Rightarrow & & U &= \int \vec E \cdot d\vec r = E d = \frac{Q d}{\varepsilon_0 (\pi - \varphi) R^2} = \frac{Q}{C} \\ \Rightarrow & & C &= \frac{\varepsilon_0 (\pi - \varphi) R^2}{d} = \frac{\varepsilon_0 A(\varphi)}{d} \end{aligned}$ with the electric field $E$ , voltage $U$ , capacitance $C$ and permittivity $\varepsilon_0$ . If a differential amount $dQ$ of charge is transferred from one capacitor surface to the other, this requires energy $dW = U dQ$ . Integration of the charge results $W = \int_{0}^Q U dQ = \int_{0}^Q \frac{Q}{C} dQ = \frac{1}{2} \frac{Q^2}{C} = \frac{d Q^2}{2 \varepsilon_0 (\pi - \varphi) R^2}$ Therefore, $W(\varphi = \dfrac{\pi}{2}) = 2 W(\varphi = 0)$

Hello, it's the first time I try to write a solution. Plus, I'm not a native speaker, so please let me know if something is unclear!

Assuming the charge $Q$ is constant, we can use Gauss' Law to calculate the field $E$ inside the capacitor.

$\oint\vec E \cdot d\vec A = EA=\frac{Q}{\varepsilon_0}$

So we get:

$E=\frac{Q}{A\varepsilon_0}$

The energy stored in the capacitor is obtained by evaluating the following integral:

$W=\frac{\varepsilon_0}{2}\iiint_V E^2 \,dV$

Where V is the volume occupied by the field. Since $E$ is constant, we get:

$W=\frac{Q^2 V}{2A^2 \varepsilon_0}$

Now, we don't even need to calculate $A$ and $V$ , since after the rotation they are both halved. The fileds are:

$E(\varphi=0°)=\frac{Q}{A\varepsilon_0}$

$E(\varphi=90°)=\frac{2Q}{A\varepsilon_0}$

And the energy stored is:

$W(\varphi=0°)=\frac{Q^2 V}{2A^2 \varepsilon_0}$

$W(\varphi=90°)=\frac{4Q^2}{2A^2 \varepsilon_0}\frac{V}{2}$

If we calculate the ratio between $W(\varphi=90°)$ and $W(\varphi=0°)$ we get 2.

I hope I didn't make mistakes.

The electrons and holes are originally a little upset about their situation, and how crowded they are, but as long as the plates are close together, they're not too upset, because they can at least see their friends. When the plates are separated, though, they become even more upset about their situation, so the energy increases.

Can anybody tell me if my solution is right or if I'm missing something?

The potential energy U stored by a capacitor is $U=\frac{1}{2}CV^2$ and since $q=CV$ then $U=\frac{1}{2}\frac{q^2}{C}$ , where C is the capacitance, q is the electrical charge and V is the potential.

During the rotation the charge remains unchanged while the capacitance is halved since it is proportional to the conductors overlapping area, so the energy doubles.