A series combination of three capacitors of capacities `1 muF, 2muF` and `8 muF` is connected to a battery of e.m.f. 13 volt .The potential difference across the plates of `2 muF` capacitor will be

Options

A. 1 V
B. 8 V
C. 4 V
D. `(13)/(3) V`

Answer

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

### Step 1: Calculate the equivalent capacitance of the series combination.
For capacitors in series, the formula to find the equivalent capacitance \( C_{eq} \) is given by:

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

Where:
- \( C_1 = 1 \mu F \)
- \( C_2 = 2 \mu F \)
- \( C_3 = 8 \mu F \)

Substituting the values:

\[ \frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{8} \]

Finding a common denominator (which is 8):

\[ \frac{1}{C_{eq}} = \frac{8}{8} + \frac{4}{8} + \frac{1}{8} = \frac{13}{8} \]

Now, taking the reciprocal to find \( C_{eq} \):

\[ C_{eq} = \frac{8}{13} \mu F \]

### Step 2: Calculate the charge on the capacitors.
The total charge \( Q \) on the capacitors in series can be calculated using the formula:

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

Where \( V = 13 \) volts (the e.m.f. of the battery). Substituting the values:

\[ Q = \left(\frac{8}{13} \mu F\right) \times 13 V = 8 \mu C \]

### Step 3: Calculate the potential difference across the \( 2 \mu F \) capacitor.
The potential difference \( V_2 \) across the \( 2 \mu F \) capacitor can be calculated using the formula:

\[ V_2 = \frac{Q}{C_2} \]

Where \( C_2 = 2 \mu F \) and \( Q = 8 \mu C \). Substituting the values:

\[ V_2 = \frac{8 \mu C}{2 \mu F} = 4 V \]

### Conclusion
The potential difference across the plates of the \( 2 \mu F \) capacitor is \( 4 \) volts.

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