The plates of parallel capaitor are given charges `+4Q` and `-2Q`. The capacitor is then connected across an uncharged capacitor of same capacitances first one `(=C)`. Find the final potential difference between the plates of the first capacitor.

Explanation

To solve the problem, we need to find the final potential difference between the plates of the first capacitor after it is connected to an uncharged capacitor of the same capacitance. Let's break this down step by step.

### Step 1: Understand the Initial Charges
The first capacitor has charges of +4Q and -2Q on its plates.

### Step 2: Calculate the Initial Charge on the Capacitor
The net charge on the first capacitor can be calculated as follows:
- The charge on one plate is +4Q.
- The charge on the other plate is -2Q.
- The effective charge (Q_eff) is given by:
\[ Q_{\text{eff}} = +4Q - 2Q = +2Q \]

### Step 3: Determine the Charge Distribution
When the first capacitor is connected to the uncharged capacitor, charge will redistribute between the two capacitors. The total charge before connection is +2Q.

### Step 4: Calculate the Final Charges
Let \( Q_1' \) and \( Q_2' \) be the final charges on the plates of the first capacitor after connection, and since the second capacitor is uncharged initially, it will take some charge from the first capacitor.

Using charge conservation:
\[ Q_1' + Q_2' = 2Q \]
Since the second capacitor is uncharged, it will take a portion of the charge from the first capacitor.

### Step 5: Set Up the Equations
Let the final potential difference across the first capacitor be \( V \). The charge on the first capacitor can be expressed as:
\[ Q_1' = C \cdot V \]
For the second capacitor, which is also of capacitance \( C \), the charge will be:
\[ Q_2' = C \cdot V_2 \]
Since the second capacitor starts with 0 charge, we can assume:
\[ Q_2' = -Q_1' + 2Q \]

### Step 6: Solve for the Final Potential
From the charge conservation equation:
\[ Q_1' + Q_2' = 2Q \]
Substituting \( Q_2' \):
\[ Q_1' - Q_1' + 2Q = 2Q \]
This means:
\[ Q_1' = 3Q/2 \]

### Step 7: Calculate the Final Potential Difference
Now, substituting back to find the potential difference:
\[ V = \frac{Q_1'}{C} = \frac{3Q/2}{C} = \frac{3Q}{2C} \]

### Final Answer
The final potential difference between the plates of the first capacitor is:
\[ V = \frac{3Q}{2C} \]