Three capacitors of `2 muF, 3 muF` and `6 muF` are joined in series and the combination is charged by means of a `24` volt battery. The potential difference between the plates of the `6 muF` capacitor is

Options

A. 4 V
B. 6 V
C. 8 V
D. 10 V

Answer

To solve the problem step by step, we will follow the process of finding the equivalent capacitance, the charge on the capacitors, and then the potential difference across the 6 µF capacitor.

### Step 1: Calculate the Equivalent Capacitance
For capacitors in series, the formula for equivalent capacitance \( C_{eq} \) is given by:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

Where:
- \( C_1 = 2 \, \mu F \)
- \( C_2 = 3 \, \mu F \)
- \( C_3 = 6 \, \mu F \)

Substituting the values:

\[ \frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \]

Finding a common denominator (which is 6):

\[ \frac{1}{C_{eq}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1 \]

Thus,

\[ C_{eq} = 1 \, \mu F \]

### Step 2: Calculate the Charge on the Capacitors
The charge \( Q \) on the capacitors in series is the same and can be calculated using the equivalent capacitance and the total voltage \( V \):

\[ Q = C_{eq} \times V \]

Where \( V = 24 \, V \):

\[ Q = 1 \, \mu F \times 24 \, V = 24 \, \mu C \]

### Step 3: Calculate the Potential Difference Across the 6 µF Capacitor
The potential difference \( V_3 \) across a capacitor is given by:

\[ V = \frac{Q}{C} \]

For the 6 µF capacitor:

\[ V_3 = \frac{Q}{C_3} = \frac{24 \, \mu C}{6 \, \mu F} \]

Calculating this gives:

\[ V_3 = \frac{24}{6} = 4 \, V \]

### Final Answer
The potential difference between the plates of the 6 µF capacitor is **4 volts**.

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