A parallel plate capacitor has a capacity `80 xx 10^(-6) F` when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of 30 V by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

Options

A. `45.6 xx 10^(-3) C`
B. `25.3 xx 10^(-3) C `
C. `120 xx 10^(-3)C`
D. `12 xx 10^(-3) C`

Answer

To solve the problem step by step, we will follow the sequence of events described in the question and apply the relevant formulas.

### Step 1: Identify the initial capacitance
The initial capacitance \( C \) of the capacitor when air is present is given as:
\[ C = 80 \times 10^{-6} \, \text{F} \]

### Step 2: Determine the capacitance with the dielectric
When a dielectric slab with a dielectric constant \( k = 20 \) is inserted, the capacitance \( C' \) of the capacitor increases. The formula for the capacitance with a dielectric is:
\[ C' = k \cdot C \]
Substituting the values:
\[ C' = 20 \cdot (80 \times 10^{-6}) = 1600 \times 10^{-6} \, \text{F} = 1.6 \times 10^{-3} \, \text{F} \]

### Step 3: Connect the capacitor to the battery
The capacitor is then connected to a battery of \( V = 30 \, \text{V} \). The charge \( Q' \) stored in the capacitor with the dielectric slab is given by:
\[ Q' = C' \cdot V \]
Substituting the values:
\[ Q' = (1.6 \times 10^{-3}) \cdot 30 = 48 \times 10^{-3} \, \text{C} = 0.048 \, \text{C} \]

### Step 4: Remove the dielectric slab
When the dielectric slab is removed, the capacitance returns to its original value \( C \). The charge \( Q \) on the capacitor after removing the dielectric is given by:
\[ Q = C \cdot V \]
Substituting the values:
\[ Q = (80 \times 10^{-6}) \cdot 30 = 2.4 \times 10^{-3} \, \text{C} = 0.0024 \, \text{C} \]

### Step 5: Calculate the charge that passes through the wire
The charge that passes through the wire when the dielectric is removed is the difference between the charge with the dielectric and the charge without it:
\[ \Delta Q = Q' - Q \]
Substituting the values:
\[ \Delta Q = 0.048 \, \text{C} - 0.0024 \, \text{C} = 0.0456 \, \text{C} \]

### Final Answer
The charge that passes through the wire when the dielectric slab is removed is:
\[ \Delta Q = 45.6 \times 10^{-3} \, \text{C} \]