A parallel plate capicitaor with plates of area `1m^2` each, are at a separation of `0.1`m. If the electric field between the plates is 100 `N//C`, the magnitude of charge on each plate is : (Take `epsilon=8.85xx10^-12(C^2)/(N-m^2)`)

Options

A. `7.85xx10^(-10)C`
B. `6.85xx10^(-10)C`
C. `8.85xx10^(-10)C`
D. `9.85xx10^(-10)C`

Explanation

To find the magnitude of charge on each plate of the parallel plate capacitor, we can follow these steps:

### Step 1: Understand the relationship between electric field and charge density
The electric field \( E \) between the plates of a parallel plate capacitor is given by the formula:
\[ E = \frac{\sigma}{\epsilon_0} \]
where:
- \( E \) is the electric field,
- \( \sigma \) is the surface charge density (charge per unit area),
- \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Relate charge density to charge
The surface charge density \( \sigma \) can be expressed in terms of the total charge \( Q \) on the plates and the area \( A \) of the plates:
\[ \sigma = \frac{Q}{A} \]

### Step 3: Substitute the expression for charge density into the electric field equation
Substituting \( \sigma \) into the electric field equation gives:
\[ E = \frac{Q/A}{\epsilon_0} \]

### Step 4: Rearrange the equation to solve for charge \( Q \)
Rearranging the equation to solve for \( Q \):
\[ Q = E \cdot A \cdot \epsilon_0 \]

### Step 5: Substitute the known values
Now we can substitute the known values into the equation:
- Area \( A = 1 \, m^2 \)
- Electric field \( E = 100 \, N/C \)
- Permittivity \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \)

So,
\[ Q = 100 \, N/C \cdot 1 \, m^2 \cdot 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \]

### Step 6: Calculate the charge \( Q \)
Calculating the above expression:
\[ Q = 100 \cdot 1 \cdot 8.85 \times 10^{-12} \]
\[ Q = 8.85 \times 10^{-10} \, C \]

### Conclusion
The magnitude of charge on each plate is:
\[ Q \approx 8.85 \times 10^{-10} \, C \]