Two capacitor of capacitance `2muF and 3muF` respectively are charged to potential difference 20 V each. Now the capacitors are connected such that the positively chargd plate of one capacitory is connected to the negatively charged plate of the other. after the current in the circuit has become negligible, the potential difference across the `2muF` capacitor is ____V.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the initial charges on each capacitor
The charge on a capacitor is given by the formula:
\[ Q = C \times V \]

For the 2 µF capacitor:
\[ Q_1 = C_1 \times V_1 = 2 \, \mu F \times 20 \, V = 40 \, \mu C \]

For the 3 µF capacitor:
\[ Q_2 = C_2 \times V_2 = 3 \, \mu F \times 20 \, V = 60 \, \mu C \]

### Step 2: Determine the net charge when connected
When the capacitors are connected with the positive plate of one connected to the negative plate of the other, the net charge on the combined system is:
\[ Q_{net} = Q_2 - Q_1 = 60 \, \mu C - 40 \, \mu C = 20 \, \mu C \]

### Step 3: Calculate the equivalent capacitance
When capacitors are connected in this manner (one plate of each connected), they effectively act in parallel. The equivalent capacitance \( C_{eq} \) is given by:
\[ C_{eq} = C_1 + C_2 = 2 \, \mu F + 3 \, \mu F = 5 \, \mu F \]

### Step 4: Calculate the potential difference across the capacitors
The potential difference \( V \) across the equivalent capacitance can be calculated using the formula:
\[ V = \frac{Q_{net}}{C_{eq}} \]

Substituting the values:
\[ V = \frac{20 \, \mu C}{5 \, \mu F} = 4 \, V \]

### Final Answer:
The potential difference across the 2 µF capacitor is **4 V**.

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