Two capacitors, `3mu F` and `4mu F`, are individually charged across a 6V battery. After being disconnected from the battery, they are connected together with the negative place of one attached to the positive plate of the other. What is the common potential?

Options

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

Answer

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

### 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 3 µF capacitor charged at 6V:
\[ Q_1 = 3 \, \mu F \times 6 \, V = 18 \, \mu C \]

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

### Step 2: Determine the net charge after connecting the capacitors
When the two capacitors are connected with opposite polarities (negative plate of one to positive plate of the other), the net charge \( Q_{net} \) is given by:
\[ Q_{net} = Q_2 - Q_1 = 24 \, \mu C - 18 \, \mu C = 6 \, \mu C \]

### Step 3: Calculate the equivalent capacitance
When the capacitors are connected in this manner, they can be treated as being in parallel. The equivalent capacitance \( C_{eq} \) is the sum of the individual capacitances:
\[ C_{eq} = C_1 + C_2 = 3 \, \mu F + 4 \, \mu F = 7 \, \mu F \]

### Step 4: Calculate the common potential
The common potential \( V_{common} \) across the capacitors can be calculated using the formula:
\[ V_{common} = \frac{Q_{net}}{C_{eq}} \]
Substituting the values we calculated:
\[ V_{common} = \frac{6 \, \mu C}{7 \, \mu F} = \frac{6}{7} \, V \approx 0.857 \, V \]

### Final Answer
The common potential after connecting the two capacitors is approximately \( 0.857 \, V \).

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