In a capacitor of capacitance `20 muF`,the distance between the plates is `2 mm`. If a dielectric slab of width `1 mm` and dielectric constant `12` is inserted between the plates, then the new capacitance is

Options

A. `2 muF`
B. `15.5 muF`
C. `26.6 muF`
D. `32 muF`

Explanation

To find the new capacitance of the capacitor after inserting a dielectric slab, we can follow these steps:

### Step 1: Understand the Configuration
The original capacitor has a capacitance \( C_0 = 20 \, \mu F \) and a distance between the plates \( d = 2 \, mm \). A dielectric slab of width \( d_1 = 1 \, mm \) and dielectric constant \( K = 12 \) is inserted between the plates.

### Step 2: Calculate the Capacitance of Each Section
When the dielectric slab is inserted, the capacitor can be viewed as two capacitors in series:
1. Capacitor 1 (with the dielectric slab):
- Width = \( d_1 = 1 \, mm \)
- Capacitance \( C_1 = \frac{K \cdot \epsilon_0 \cdot A}{d_1} \)

2. Capacitor 2 (without the dielectric slab):
- Width = \( d_2 = d - d_1 = 2 \, mm - 1 \, mm = 1 \, mm \)
- Capacitance \( C_2 = \frac{\epsilon_0 \cdot A}{d_2} \)

### Step 3: Find the Area \( A \)
The capacitance formula is given by:
\[ C = \frac{\epsilon_0 \cdot A}{d} \]
From the original capacitance:
\[ C_0 = \frac{\epsilon_0 \cdot A}{d} \]
Substituting the known values:
\[ 20 \times 10^{-6} = \frac{\epsilon_0 \cdot A}{2 \times 10^{-3}} \]
From this, we can find \( \epsilon_0 \cdot A \):
\[ \epsilon_0 \cdot A = 20 \times 10^{-6} \times 2 \times 10^{-3} = 40 \times 10^{-9} \, F \cdot m \]

### Step 4: Calculate \( C_1 \) and \( C_2 \)
Now we can calculate \( C_1 \) and \( C_2 \):
1. For \( C_1 \):
\[ C_1 = \frac{K \cdot \epsilon_0 \cdot A}{d_1} = \frac{12 \cdot (40 \times 10^{-9})}{1 \times 10^{-3}} = 480 \times 10^{-6} \, F = 480 \, \mu F \]

2. For \( C_2 \):
\[ C_2 = \frac{\epsilon_0 \cdot A}{d_2} = \frac{40 \times 10^{-9}}{1 \times 10^{-3}} = 40 \times 10^{-6} \, F = 40 \, \mu F \]

### Step 5: Calculate the Total Capacitance
Since \( C_1 \) and \( C_2 \) are in series, the total capacitance \( C \) can be calculated using:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C} = \frac{1}{480 \times 10^{-6}} + \frac{1}{40 \times 10^{-6}} \]
Calculating:
\[ \frac{1}{C} = \frac{1}{480} + \frac{1}{40} \]
Finding a common denominator (480):
\[ \frac{1}{C} = \frac{1}{480} + \frac{12}{480} = \frac{13}{480} \]
Thus,
\[ C = \frac{480}{13} \approx 36.92 \, \mu F \]

### Final Answer
The new capacitance after inserting the dielectric slab is approximately \( 36.92 \, \mu F \).

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