An air filled parallel plate capacitor of capacitance `50 mu F` is connected to a battery of 100 V. A slab of dielectric constant 4 is inserted in it to fill the space completely . Find the extra charge flown through the battery till it attains the steady state.

Options

A. `2.5 mC`
B. `5 mC`
C. `20 mC`
D. ` 15 mC`

Answer

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

### Step 1: Calculate the initial charge on the capacitor without the dielectric.
The formula for the charge \( Q \) on a capacitor is given by:
\[ Q = C \cdot V \]
where \( C \) is the capacitance and \( V \) is the voltage.

Given:
- Initial capacitance \( C_0 = 50 \, \mu F = 50 \times 10^{-6} \, F \)
- Voltage \( V_0 = 100 \, V \)

Now, substituting the values:
\[ Q_0 = C_0 \cdot V_0 = 50 \times 10^{-6} \, F \cdot 100 \, V = 5000 \times 10^{-6} \, C = 5 \, mC \]

### Step 2: Calculate the new capacitance when the dielectric is inserted.
When a dielectric is inserted, the capacitance increases by a factor equal to the dielectric constant \( K \).

Given:
- Dielectric constant \( K = 4 \)

The new capacitance \( C' \) is given by:
\[ C' = K \cdot C_0 = 4 \cdot 50 \, \mu F = 200 \, \mu F = 200 \times 10^{-6} \, F \]

### Step 3: Calculate the new charge on the capacitor with the dielectric.
Using the same formula for charge:
\[ Q' = C' \cdot V \]

Substituting the values:
\[ Q' = 200 \times 10^{-6} \, F \cdot 100 \, V = 20000 \times 10^{-6} \, C = 20 \, mC \]

### Step 4: Calculate the extra charge that flowed through the battery.
The extra charge \( \Delta Q \) that flowed through the battery is the difference between the new charge and the initial charge:
\[ \Delta Q = Q' - Q_0 \]

Substituting the values:
\[ \Delta Q = 20 \, mC - 5 \, mC = 15 \, mC \]

### Final Answer:
The extra charge that flowed through the battery till it attains the steady state is \( 15 \, mC \).

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