A parallel plate capacitor of capacity `5 mu F` and plate separation `6 cm` is connected to a `1V` battery and is charged. A dielectric of dielectric constant `4` and thickness `4 cm` is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is.

Options

A. `2 muC`
B. `3 muC`
C. `5 mu C`
D. `10 muC`

Explanation

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

### Step 1: Calculate the initial charge on the capacitor
Given:
- Capacitance \( C = 5 \, \mu F = 5 \times 10^{-6} \, F \)
- Voltage \( V = 1 \, V \)

The charge \( Q \) on the capacitor can be calculated using the formula:
\[ Q = C \times V \]
Substituting the values:
\[ Q = 5 \times 10^{-6} \, F \times 1 \, V = 5 \times 10^{-6} \, C = 5 \, \mu C \]

### Step 2: Determine the new capacitance after introducing the dielectric
The dielectric has:
- Dielectric constant \( K = 4 \)
- Thickness \( t = 4 \, cm = 0.04 \, m \)
- Plate separation \( D = 6 \, cm = 0.06 \, m \)

The new capacitance \( C' \) can be calculated using the formula for a capacitor with a dielectric:
\[ C' = \frac{A \epsilon_0}{D - t + \frac{t}{K}} \]
However, we can also express the new capacitance in terms of the original capacitance:
\[ C' = C \cdot \frac{1}{1 - \frac{t}{D} + \frac{t}{K \cdot D}} \]
Substituting the values:
\[ C' = 5 \, \mu F \cdot \frac{1}{1 - \frac{0.04}{0.06} + \frac{0.04}{4 \cdot 0.06}} \]

Calculating the fractions:
\[ \frac{0.04}{0.06} = \frac{2}{3} \quad \text{and} \quad \frac{0.04}{4 \cdot 0.06} = \frac{0.04}{0.24} = \frac{1}{6} \]
Thus:
\[ C' = 5 \, \mu F \cdot \frac{1}{1 - \frac{2}{3} + \frac{1}{6}} = 5 \, \mu F \cdot \frac{1}{\frac{1}{3} + \frac{1}{6}} = 5 \, \mu F \cdot \frac{1}{\frac{2}{6} + \frac{1}{6}} = 5 \, \mu F \cdot \frac{1}{\frac{3}{6}} = 5 \, \mu F \cdot \frac{2}{3} \]
Calculating \( C' \):
\[ C' = 5 \, \mu F \cdot \frac{2}{3} = \frac{10}{3} \, \mu F \approx 10 \, \mu F \]

### Step 3: Calculate the new charge on the capacitor
Now, we can calculate the new charge \( Q' \) using the new capacitance:
\[ Q' = C' \times V = 10 \, \mu F \times 1 \, V = 10 \, \mu C \]

### Step 4: Calculate the additional charge that flows into the capacitor
The additional charge \( \Delta Q \) that flows into the capacitor is given by:
\[ \Delta Q = Q' - Q = 10 \, \mu C - 5 \, \mu C = 5 \, \mu C \]

### Final Answer
The additional charge that flows into the capacitor from the battery is \( 5 \, \mu C \).

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