Question

Find the charge on $C$ as a function of time.

Answer 1

Hint:The initial states of both the capacitors are given to us in the question, now, at steady state, consider a charge ${q_f}$ at the capacitor $C$, the remaining charge would be at the capacitor $2C$. Now, apply the loop law to find the final value of the charge ${q_f}$. Now that you’ve found out the initial and final values of charge at the capacitor $C$, you can use the modified formula which gives the equation of charge on a capacitor when the capacitor is being discharged as following:
$q = {q_i} + \left( {{q_f} - {q_i}} \right)\left( {1 - {e^{ - \dfrac{t}{\tau }}}} \right)$

Complete step by step answer:
We will proceed exactly as we described in the hint section of the solution to the question.
Initially, the charge at $C$ and $2C$ is given to us as ${q_i} = \dfrac{{CE}}{2}$ and $0$
Let’s consider the charge at $C$ at steady state be ${q_f}$. We can observe that the remaining charge $\left( {\dfrac{{CE}}{2} - {q_f}} \right)$ will be present at the capacitor $2C$ .
At the steady state, we know that no current will flow in the circuit, hence, we can write the following equation using the loop law:
$E - \dfrac{{{q_f}}}{C} + \dfrac{{\dfrac{{CE}}{2} - {q_f}}}{{2C}} = 0$
Upon solving the equation, we get:
${q_f} = \dfrac{5}{6}CE$
We have found out that the charge on capacitor $C$ has dropped from an initial value of ${q_i} = \dfrac{{CE}}{2}$ to a steady state or final value of ${q_f} = \dfrac{5}{6}CE$
Now, we need to find the value of time constant $\tau $, which can be found out using the formula:
$\tau = {C_{net}}{R_{net}}$
Where, ${C_{net}}$ is the net capacitance in the circuit and,
${R_{net}}$ is the net resistance in the circuit.
Let’s now find the value of net capacitance of the circuit. We can clearly see that the capacitors are in series arrangement, hence the net capacitance can be found out using the formula:
$\dfrac{1}{{{C_{net}}}} = \sum {\dfrac{1}{{{C_i}}}} $
Applying the above-mentioned formula, we get:
$\dfrac{1}{{{C_{net}}}} = \dfrac{1}{C} + \dfrac{1}{{2C}}$
Upon solving, we get:
${C_{net}} = \dfrac{2}{3}C$
We can clearly see that the net resistance of the circuit is simply ${R_{net}} = R$
Now, our time constant comes out as:
$\tau = \dfrac{2}{3}CR$
Substituting the values that we found out in the above-mentioned formula:
$q = {q_i} + \left( {{q_f} - {q_i}} \right)\left( {1 - {e^{ - \dfrac{t}{\tau }}}} \right)$
We get:
$q = \dfrac{{CE}}{2} + \left( {\dfrac{5}{6}CE - \dfrac{{CE}}{2}} \right)\left( {1 - {e^{ - \dfrac{{3t}}{{2CR}}}}} \right)$
Upon solving, we get:
$q = \dfrac{{CE}}{2} + \dfrac{{CE}}{3}\left( {1 - {e^{ - \dfrac{{3t}}{{2CR}}}}} \right)$
This is the final answer to the question asked.

Note:This is a faster way to solve the question, the more basic one is slightly tougher and lengthier than the method that we used to solve the question. In that method, you apply the loop law at an intermediate state when the current flows in the circuit with a value of $i$ and write it as $i = \dfrac{{dq}}{{dt}}$ and integrate both sides to find the value of charge as a function of time.