In the above problem the value of E in the space outside the sheets is .

Options

A. `sigma //in_(0)`
B. `sigma //2in_(0)`
C. zero
D. `2sigma //in_(0)`

Answer

To solve the problem of finding the electric field \( E \) in the space outside two charged sheets, we can follow these steps:

### Step 1: Understand the Configuration
We have two infinite sheets: one with a positive surface charge density \( \sigma \) and the other with a negative surface charge density \( -\sigma \). The electric field produced by a single infinite sheet of charge is given by the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Determine the Electric Field Inside the Sheets
Inside the region between the two sheets, the electric fields due to both sheets add up because they are in the same direction (from the positive to the negative sheet). Therefore, the total electric field \( E_{\text{inside}} \) between the sheets is:
\[ E_{\text{inside}} = E_{\text{positive}} + E_{\text{negative}} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \]

### Step 3: Determine the Electric Field Outside the Sheets
Now, we need to find the electric field in the regions outside the sheets. To do this, we can apply Gauss's law.

1. **Choose a Gaussian Surface**: We can choose a Gaussian surface that is outside the sheets.
2. **Evaluate the Electric Field**: The electric field due to the positive sheet points away from the sheet, while the electric field due to the negative sheet points towards it. Therefore, outside the sheets, the electric fields due to both sheets will cancel each other out.

Thus, the electric field \( E_{\text{outside}} \) in the regions outside the sheets is:
\[ E_{\text{outside}} = E_{\text{positive}} + E_{\text{negative}} = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0 \]

### Conclusion
The electric field \( E \) in the space outside the sheets is zero.

### Final Answer
\[ E_{\text{outside}} = 0 \]

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