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## Homework Statement

Initially:At t=0 [/B]the cylindrical capacitor of capacitance [itex]c=\frac{\epsilon s}{d}[/itex] (d the distance between the 2 electrodes and s their surface; [itex]\epsilon = \epsilon(T)[/itex] is the dielectric permittivity) is discharged and we close the circuit and charge isothermally the cylindrical capacitor until it has a [itex]c*v[/itex] charge.

The current is [itex]i=\frac{dq}{dt}[/itex] and the voltage is v(t), the electric charge of the capacitor is [itex]q(t)=c*v(t)[/itex].

We have an RC circuit receiving work [itex]W[/itex] from a generator of voltage [itex]v[/itex] and energy [itex]Q_{exchanged}[/itex] from a thermal reservoir (thermostat) of [itex]T_0[/itex] Kelvin,the circuit has its entropy change[itex]\Delta S_{in}[/itex] during the charging.

We have [itex]C_q[/itex] the heat capacity for a constant charge q and [itex] \lambda= \frac{qT \epsilon'}{\epsilon c}[/itex] another coefficient in the equation :

[itex]TdS_{in}= C_q dT + \lambda dq[/itex] (1)

What is the entropy of the universe [itex]\Delta S_{univ}[/itex] equal to ?

Does [itex]C_q[/itex] depend on [itex]q[/itex] ?

## Homework Equations

[itex]\Delta S_{thermostat}=\frac{Q_{exchanged}}{T_0}[/itex]

## The Attempt at a Solution

If we integrate [itex] C_q[/itex] over

**T ,**I think

**q**does appear in the formula for [itex]C_q[/itex] but it's not a variable I think.To find [itex]\Delta S_{in}[/itex] we divide the equation (1) by T then integrate [itex] \frac{\lambda}{T}[/itex] over q from 0 to q=cv.

[itex]\Delta S_{univ}=\Delta S_{thermostat}+\Delta S_{in}[/itex]

I would like to know how to express [itex]\Delta S_{thermostat}[/itex] and it's sign. Thank you.