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)=8.8xx10^(-12) Nm^(2) C^(-2))`.
Energy supplied by the battery is.

Options

A. a. `10^(-9)J`
B. b. `5xx10^(9) J`
C. c. `50xx10^(-9) J`
D. d. `25xx10^(-9) J`

Explanation

To solve the problem step by step, we will calculate the energy supplied by the battery to the parallel plate capacitor.

### Step 1: Identify the given values
- Area of each plate, \( S = 5 \times 10^{-3} \, \text{m}^2 \)
- Distance between the plates, \( d = 8.80 \, \text{mm} = 8.80 \times 10^{-3} \, \text{m} \)
- Charge on plate A, \( q_1 = +10^{-10} \, \text{C} \)
- Charge on plate B, \( q_2 = +2 \times 10^{-10} \, \text{C} \)
- Battery emf, \( E = 10 \, \text{V} \)
- Permittivity of free space, \( \epsilon_0 = 8.8 \times 10^{-12} \, \text{N m}^2/\text{C}^2 \)

### Step 2: Calculate the capacitance of the capacitor
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\epsilon_0 \cdot S}{d} \]
Substituting the values:
\[ C = \frac{8.8 \times 10^{-12} \, \text{N m}^2/\text{C}^2 \cdot 5 \times 10^{-3} \, \text{m}^2}{8.80 \times 10^{-3} \, \text{m}} \]
Calculating this gives:
\[ C = \frac{4.4 \times 10^{-14}}{8.80 \times 10^{-3}} = 5 \times 10^{-12} \, \text{F} \]

### Step 3: Calculate the charge on the capacitor
The charge \( Q \) on the capacitor can be calculated using the formula:
\[ Q = C \cdot V \]
Substituting the values:
\[ Q = 5 \times 10^{-12} \, \text{F} \cdot 10 \, \text{V} = 5 \times 10^{-11} \, \text{C} = 50 \, \text{pC} \]

### Step 4: Calculate the energy supplied by the battery
The energy \( E \) supplied by the battery can be calculated using the formula:
\[ E = \Delta Q \cdot V \]
Where \( \Delta Q \) is the change in charge. The initial charge on plate A was \( -50 \, \text{pC} \) (since it is positively charged) and the final charge is \( +50 \, \text{pC} \). Thus:
\[ \Delta Q = 50 \, \text{pC} - (-50 \, \text{pC}) = 100 \, \text{pC} \]
Now substituting into the energy formula:
\[ E = 100 \times 10^{-12} \, \text{C} \cdot 10 \, \text{V} = 1000 \times 10^{-12} \, \text{J} = 10^{-9} \, \text{J} \]

### Final Answer
The energy supplied by the battery is:
\[ E = 10^{-9} \, \text{J} \text{ or } 1 \, \text{nJ} \]