Two large, thin plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude `16xx10^(-22)C m^(-2)`. The electric field between the plates is

Options

A. `1.8xx10^(-10)NC^(-1)`
B. `1.9xx10^(-10)NC^(-1)`
C. `1.6xx10^(-10)NC^(-1)`
D. `1.5xx10^(-10)NC^(-1)`

Explanation

To find the electric field between two large, thin parallel plates with surface charge densities of opposite signs, we can follow these steps:

### Step 1: Understand the Setup
We have two parallel plates:
- Plate 1 has a surface charge density \( \sigma = +16 \times 10^{-22} \, \text{C/m}^2 \)
- Plate 2 has a surface charge density \( \sigma = -16 \times 10^{-22} \, \text{C/m}^2 \)

### Step 2: Electric Field Due to a Single Plate
The electric field \( E \) due to a single infinite plane sheet of charge with surface charge density \( \sigma \) is given by the formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 3: Calculate the Electric Field from Each Plate
For Plate 1 (positive charge):
\[ E_1 = \frac{16 \times 10^{-22}}{2 \times 8.854 \times 10^{-12}} \]

For Plate 2 (negative charge):
\[ E_2 = \frac{16 \times 10^{-22}}{2 \times 8.854 \times 10^{-12}} \]

### Step 4: Direction of the Electric Fields
- The electric field due to Plate 1 points away from the plate (to the right).
- The electric field due to Plate 2 points towards the plate (to the right).

Since both electric fields are in the same direction between the plates, we can add them together.

### Step 5: Calculate the Total Electric Field
The total electric field \( E \) between the plates is:
\[ E = E_1 + E_2 = \frac{16 \times 10^{-22}}{2 \times 8.854 \times 10^{-12}} + \frac{16 \times 10^{-22}}{2 \times 8.854 \times 10^{-12}} = 2 \cdot \frac{16 \times 10^{-22}}{2 \times 8.854 \times 10^{-12}} \]
\[ E = \frac{16 \times 10^{-22}}{8.854 \times 10^{-12}} \]

### Step 6: Perform the Calculation
Calculating the above expression:
\[ E = \frac{16 \times 10^{-22}}{8.854 \times 10^{-12}} \approx 1.8 \times 10^{-10} \, \text{N/C} \]

### Final Answer
Thus, the electric field between the plates is approximately:
\[ E \approx 1.8 \times 10^{-10} \, \text{N/C} \]