Electric field between the plates is given by: value of sigma is =17x10^(-22)

Options

A. `1.9 xx 10 ^(-10) (N)/(C) `
B. `1. 5 xx 10 ^(-15) (N)/(C)`
C. zero
D. `17 xx 10^(-10) (N)/(C)`

Explanation

To find the electric field between two charged plates given the charge density (σ), we can follow these steps:

### Step 1: Understand the Charge Density
The charge density (σ) is given as:
\[ \sigma = 17 \times 10^{-22} \, \text{C/m}^2 \]
This represents the amount of charge per unit area on the plates.

### Step 2: Electric Field Due to a Single Infinite Plate
The electric field (E) due to an infinite plane sheet of charge with surface charge density σ is given by the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where ε₀ (epsilon naught) is the permittivity of free space, approximately equal to:
\[ \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \]

### Step 3: Calculate the Electric Field from Each Plate
Since we have two plates, one with positive charge density (σ) and the other with negative charge density (-σ), the total electric field between the plates (E_net) is the sum of the electric fields due to each plate. The direction of the electric fields from both plates will be the same between the plates.

Thus, the total electric field is:
\[ E_{\text{net}} = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \]

### Step 4: Substitute the Values
Now, substituting the values of σ and ε₀ into the equation:
\[ E_{\text{net}} = \frac{17 \times 10^{-22}}{8.85 \times 10^{-12}} \]

### Step 5: Perform the Calculation
Calculating the above expression:
1. Divide the charge density by the permittivity:
\[ E_{\text{net}} = \frac{17}{8.85} \times 10^{-22 + 12} = \frac{17}{8.85} \times 10^{-10} \]
2. Calculate \( \frac{17}{8.85} \):
\[ \frac{17}{8.85} \approx 1.92 \]
3. Therefore:
\[ E_{\text{net}} \approx 1.92 \times 10^{-10} \, \text{N/C} \]

### Final Answer
The electric field between the plates is:
\[ E \approx 1.92 \times 10^{-10} \, \text{N/C} \]

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