An infinite nonconducting sheet of charge has a surface charge density of `10^(7) C//m^(2)`. The separation between two equipotential surfaces near the whose potential differ by 5V is

Options

A. 0.88 cm
B. 0.88 mm
C. 0.88 m
D. `5 xx 10^(-7)m`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given information
We are given:
- Surface charge density, \( \sigma = 10^7 \, \text{C/m}^2 \)
- Potential difference, \( V = 5 \, \text{V} \)

### Step 2: Calculate the electric field due to the infinite non-conducting sheet
The electric field \( E \) due to an infinite sheet of charge is given by the formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 3: Substitute the values into the electric field formula
Substituting the value of \( \sigma \) into the formula:
\[ E = \frac{10^7}{2 \times 8.85 \times 10^{-12}} \]

### Step 4: Calculate the electric field
Calculating the above expression:
\[ E = \frac{10^7}{1.77 \times 10^{-11}} \approx 5.64 \times 10^{17} \, \text{N/C} \]

### Step 5: Relate potential difference to electric field and distance
The relationship between potential difference \( V \), electric field \( E \), and distance \( d \) is given by:
\[ V = E \cdot d \]
From this, we can express \( d \) as:
\[ d = \frac{V}{E} \]

### Step 6: Substitute the values to find distance
Substituting the values of \( V \) and \( E \):
\[ d = \frac{5}{5.64 \times 10^{17}} \]

### Step 7: Calculate the distance \( d \)
Calculating the above expression:
\[ d \approx 0.88 \times 10^{-3} \, \text{m} = 0.88 \, \text{mm} \]

### Final Answer
The separation between the two equipotential surfaces is approximately \( 0.88 \, \text{mm} \).

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