Each plate of a parallel plate air capacitor has area `S=5xx10^(-3) m^(2)` and the distance between the plates is `d=8.80 mm`.Plate A has positive charge `q_(1)=+10^(-10) C`, and plate B has charge `q_(2)=+2xx10^(-10) C`. A battery of emf `E=10 V` has its positive terminal connected to plate A and the negative terminal to plate B. (Given `epsilon_(0)=8.8xx^(12) Nm^(2) C^(-2))`.
Charge supplied by time the battery is .

Options

A. `120 pC`
B. `100 pC`
C. `60 pC`
D. `50 pC`

Explanation

To solve the problem, we need to find the charge supplied by the battery when it is connected to the parallel plate capacitor. We will follow these steps:

### Step 1: Convert the distance between the plates from mm to meters
The distance \( d \) is given as \( 8.80 \, \text{mm} \). To convert this to meters:
\[ d = 8.80 \, \text{mm} = 8.80 \times 10^{-3} \, \text{m} \]

**Hint:** Remember to convert all units to SI units for consistency.

### Step 2: Calculate the capacitance of the parallel plate capacitor
The formula for the capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 \cdot S}{d} \]
where:
- \( \epsilon_0 = 8.8 \times 10^{-12} \, \text{N m}^2/\text{C}^2 \) (permittivity of free space),
- \( S = 5 \times 10^{-3} \, \text{m}^2 \) (area of the plates),
- \( d = 8.80 \times 10^{-3} \, \text{m} \) (distance between the plates).

Substituting the values:
\[ C = \frac{8.8 \times 10^{-12} \cdot 5 \times 10^{-3}}{8.80 \times 10^{-3}} = \frac{44 \times 10^{-15}}{8.80 \times 10^{-3}} = 5 \times 10^{-12} \, \text{F} \]

**Hint:** Make sure to keep track of the units when performing calculations.

### Step 3: Calculate the charge supplied by the battery
The charge \( Q \) supplied by the battery can be calculated using the formula:
\[ Q = C \cdot V \]
where \( V = 10 \, \text{V} \) (emf of the battery).

Substituting the values:
\[ Q = 5 \times 10^{-12} \cdot 10 = 50 \times 10^{-12} \, \text{C} = 50 \, \text{pC} \]

**Hint:** Remember that 1 picoCoulomb (pC) is \( 10^{-12} \, \text{C} \).

### Final Answer
The charge supplied by the battery when it is connected to the capacitor is:
\[ Q = 50 \, \text{pC} \]

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