Block, Capacitor, and Pneumatic Cylinder

The electrical potential energy $U_{cap}$ , of a capacitance $C$ , carrying charge $Q_{cap}$ is:

$U_{cap} = \frac { {Q_{cap}}^2 } { 2 C }$

For a parallel plate capacitor, with plate area $A_{cap}$ , plate separation $d_{cap}$ , and no dielectric:

$C = \frac { \epsilon_0 A_{cap} } { d_{cap} }$

Thus,

$U_{cap} = \frac { {Q_{cap}}^2 d_{cap} } { 2 \epsilon_0 A_{cap} }$

If we let $\lambda$ be the value of $U_{cap}$ at the start of phase 1, which is also the end of phase 4, then, $4 \lambda$ , $8 \lambda$ , and $2 \lambda$ , are the corresponding values of $U_{cap}$ at the end of phases 1, 2, and 3, respectively.

The corresponding values for the change in electrical potential energy $\Delta U_{cap}$ of the capacitor, for phases 1, 2, 3, and 4, are $3 \lambda$ , $4 \lambda$ , $-6 \lambda$ , and $- \lambda$ , respectively.

Two processes that contribute to $\Delta U_{cap}$ are:

1) The electrical work done by external circuitry to transfer charge carriers between the plates, $W_{term}$

2) The mechanical work done by the electrical force on the block-right plate-piston system, $W_{EB}$

Mathematically,

$\Delta U_{cap} = W_{term} - W_{EB}$

For phases 1 and 3, the block maintains its position, $W_{EB} = 0$ , and for phases 2 and 4, the switch is open, $W_{term} = 0$ .

We can now complete this table:

There are only two forces acting on the block-right plate-piston system that have a horizontal component. In fact, both forces are purely horizontal:

1) The rightward push by the gas on the piston

2) The leftward pull by the left plate on the right plate

To achieve mechanical equilibrium, the two forces must have equal magnitudes:

$P_{gas} A_{piston} = Q_{cap} E_{left}$

$E_{left}$ is the contribution of the left plate to the electric field of the capacitor which we can determine using Gauss' Law:

$E_{left} = \frac { Q_{cap} } { 2 \epsilon_0 A_{cap} }$

From the Ideal Gas Law, where $d_{gas}$ is the distance between the left end of the piston and the left end of the enclosure, and $R$ is the ideal gas constant:

$P_{gas} A_{piston} d_{gas} = n_{gas} R T_{gas}$

The last three equations will yield:

$\frac { { Q_{cap} }^2 d_{gas} } { 2 \epsilon_0 A_{cap} } = n_{gas} R T_{gas}$

It can be shown that $d_{gas} = d_{cap}$ , and recognizing $U_{cap}$ , we get:

$U_{cap} = n_{gas} R T_{gas}$

The internal energy of an ideal monatomic gas is:

$U_{gas} = \frac {3} {2} n_{gas} R T_{gas}$

Thus,

$U_{gas} = \frac {3} {2} U_{cap}$

$\Delta U_{gas} = \frac {3} {2} \Delta U_{cap}$

The two forces cancel each other, so the work done by the gas must be:

$W_{gas} = - W_{EB}$

The heat transferred to the gas can be calculated using the First Law of Thermodynamics:

$\Delta U_{gas} = Q_{gas} - W_{gas}$

We apply the last three equations on the entries of the first table, to calculate the entries of this second table:

In Phase 3, the capacitor delivers electrical energy to the electrical load:

$W_{LOAD} = 6 \lambda$

In Phase 1, the electrical source supplies electrical energy to the capacitor:

$W_{SOURCE} = 3 \lambda$

In Phases 1 and 2, the gas absorbs heat from the heat source:

$Q_{SOURCE} = \frac {29} {2} \lambda$

Therefore, the efficiency of the cycle is:

$\frac { W_{LOAD} } { W_{SOURCE} + Q_{SOURCE} } = \boxed {\frac {12} {35}}$