How do you set up Kirchhoff's loop rule for a circuit with a charged capacitor, uncharged capacitor, and resistor?

Question

If I connect a circuit with charged capacitor that has charge $q$ to an uncharged capacitor and resistor, how would I set up a differential equation that can be used to find the time constant? I know that at any time $q_1 + q_2 = q$ where $q_1$ is the charge on the initially charged capacitor and $q_2$ is the charge on the initially uncharged capacitor. I thought that $(frac{q_1}{C_1})-(frac{q-q_1}{C_2})-R(-frac{dq_1}{dt})=0$ might work, but I am not sure that this correctly describes the circuit. Are the signs on the second capacitor and resistor supposed to match? Is $frac{dq_1}{dt}$ representative of the circuit current? Once I have that differential equation made, how can I obtain the time constant by comparing to the basic RC-circuit $frac{dq}{dt}=frac{CV-q}{RC}$?

DanDan0101 · Accepted Answer

Your diff. eq. looks perfectly fine, based on your sign conventions for $q_1$ and $q_2$. Note that $frac{mathrm{d}q_1}{mathrm{d}t}=-frac{mathrm{d}q_2}{mathrm{d}t}$ under this convention. Both of these are representative of the circuit current, just with different signs, since everything is connected in series.
Let's define $Cequivfrac{1}{frac{1}{C_1}+frac{1}{C_2}}$ as the equivalent capacitance of the two capacitors; and $Vequivfrac{q}{C_2}$ as a sort of voltage constant. Then, the diff. eq. takes the form:
$$frac{mathrm{d}q_1}{mathrm{d}t}=frac{CV-q_1}{RC}$$
So, as usual, we can find the time constant:
$$tau=RC=frac{R}{frac{1}{C_1}+frac{1}{C_2}}$$

How do you set up Kirchhoff's loop rule for a circuit with a charged capacitor, uncharged capacitor, and resistor?

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