Two capcitors, 3 `muF` and `4 muF,` are individually charged across a 6 V battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the final total energy stored ?

Options

A. `1.26xx10^(-4)J`
B. `2.57xx10^(-4)J`
C. `1.25xx10^(-6)J`
D. `2.57xx10^(-6)J`

Explanation

To solve the problem step by step, we will follow the procedure outlined in the video transcript.

### Step 1: Calculate the charge on each capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where \( C \) is the capacitance and \( V \) is the voltage.

For the first capacitor \( C_1 = 3 \, \mu F \) charged at \( V = 6 \, V \):
\[ Q_1 = C_1 \times V = 3 \, \mu F \times 6 \, V = 18 \, \mu C \]

For the second capacitor \( C_2 = 4 \, \mu F \) charged at \( V = 6 \, V \):
\[ Q_2 = C_2 \times V = 4 \, \mu F \times 6 \, V = 24 \, \mu C \]

### Step 2: Determine the configuration after connecting the capacitors
When the capacitors are disconnected from the battery and connected together with the negative plate of one attached to the positive plate of the other, the total charge \( Q_{total} \) is:
\[ Q_{total} = Q_2 - Q_1 = 24 \, \mu C - 18 \, \mu C = 6 \, \mu C \]

### Step 3: Calculate the total capacitance
The total capacitance \( C_{total} \) when capacitors are connected in series is given by:
\[ C_{total} = C_1 + C_2 = 3 \, \mu F + 4 \, \mu F = 7 \, \mu F \]

### Step 4: Calculate the common potential
The common potential \( V \) across the capacitors can be calculated using:
\[ V = \frac{Q_{total}}{C_{total}} = \frac{6 \, \mu C}{7 \, \mu F} = \frac{6}{7} \, V \]

### Step 5: Calculate the total energy stored
The energy \( U \) stored in the capacitors is given by the formula:
\[ U = \frac{1}{2} C_{total} V^2 \]
Substituting the values:
\[ U = \frac{1}{2} \times 7 \times 10^{-6} \, F \times \left(\frac{6}{7} \, V\right)^2 \]
Calculating \( V^2 \):
\[ \left(\frac{6}{7}\right)^2 = \frac{36}{49} \]
Now substituting back into the energy equation:
\[ U = \frac{1}{2} \times 7 \times 10^{-6} \times \frac{36}{49} \]
\[ U = \frac{1}{2} \times \frac{252 \times 10^{-6}}{49} \]
\[ U = \frac{126 \times 10^{-6}}{49} \approx 2.57 \times 10^{-6} \, J \]

### Final Answer
The final total energy stored is approximately:
\[ U \approx 2.57 \times 10^{-6} \, J \]

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