A capacitor of capacitance `C` is given a charge q0. At time `t = 0` it is connected to an uncharged capacitor of equal capacitance through a resistance `R`. Find the charge on the first capacitor and the second capacitor as a function of time `t`. Also plot the corresponding q-t graphs.

Explanation

To solve the problem step by step, we will analyze the circuit involving two capacitors and a resistor, applying Kirchhoff's laws.

### Step 1: Understand the Initial Conditions
At time \( t = 0 \):
- Capacitor 1 has charge \( q_0 \).
- Capacitor 2 is uncharged, so it has charge \( 0 \).
- The two capacitors are connected through a resistor \( R \).

### Step 2: Define the Charges on the Capacitors
Let:
- \( q \) be the charge on Capacitor 2 at time \( t \).
- The charge on Capacitor 1 at time \( t \) will then be \( q_0 - q \).

### Step 3: Apply Kirchhoff's Law
According to Kirchhoff's law, the sum of the potential differences in the loop must equal zero. The potential difference across Capacitor 1, the resistor, and Capacitor 2 can be expressed as:
\[ \frac{q_0 - q}{C} + I R - \frac{q}{C} = 0 \]
where \( I \) is the current flowing through the circuit.

### Step 4: Express Current in Terms of Charge
The current \( I \) can be expressed as:
\[ I = -\frac{dq}{dt} \]
Substituting this into the Kirchhoff's equation gives:
\[ \frac{q_0 - q}{C} - R \frac{dq}{dt} - \frac{q}{C} = 0 \]
This simplifies to:
\[ \frac{q_0 - 2q}{C} = R \frac{dq}{dt} \]

### Step 5: Rearrange the Equation
Rearranging the equation, we have:
\[ R \frac{dq}{dt} = \frac{q_0 - 2q}{C} \]
or
\[ \frac{dq}{q_0 - 2q} = \frac{dt}{RC} \]

### Step 6: Integrate Both Sides
Integrating both sides:
\[ \int \frac{dq}{q_0 - 2q} = \int \frac{dt}{RC} \]
The left side integrates to:
\[ -\frac{1}{2} \ln |q_0 - 2q| \quad \text{(using the substitution \( u = q_0 - 2q \))} \]
The right side integrates to:
\[ \frac{t}{RC} + C_1 \]

### Step 7: Solve for the Constant of Integration
At \( t = 0 \), \( q = 0 \):
\[ -\frac{1}{2} \ln |q_0| = C_1 \]
Thus, we can express the equation as:
\[ -\frac{1}{2} \ln |q_0 - 2q| = \frac{t}{RC} - \frac{1}{2} \ln |q_0| \]

### Step 8: Exponentiate to Solve for \( q \)
Exponentiating both sides gives:
\[ |q_0 - 2q| = q_0 e^{-\frac{2t}{RC}} \]
This leads to:
\[ q_0 - 2q = q_0 e^{-\frac{2t}{RC}} \]
Solving for \( q \):
\[ 2q = q_0 - q_0 e^{-\frac{2t}{RC}} \]
\[ q = \frac{q_0}{2} (1 - e^{-\frac{2t}{RC}}) \]

### Step 9: Find Charge on Capacitor 1
The charge on Capacitor 1 is:
\[ q_1 = q_0 - q = q_0 - \frac{q_0}{2} (1 - e^{-\frac{2t}{RC}}) = \frac{q_0}{2} (1 + e^{-\frac{2t}{RC}}) \]

### Step 10: Summary of Results
The charges on the capacitors as functions of time are:
- Charge on Capacitor 2:
\[ q(t) = \frac{q_0}{2} (1 - e^{-\frac{2t}{RC}}) \]
- Charge on Capacitor 1:
\[ q_1(t) = \frac{q_0}{2} (1 + e^{-\frac{2t}{RC}}) \]

### Step 11: Plotting the Graphs
- The charge on Capacitor 1 starts at \( q_0 \) and decreases to \( \frac{q_0}{2} \).
- The charge on Capacitor 2 starts at \( 0 \) and increases to \( \frac{q_0}{2} \).