An infinite sheet of charge has a surfaces charge density of `10^-7C/m^2`. The separation between two equipotential surfce whose potentials differ by `5 V` is

Options

A. `0.64cm`
B. `0.88mm`
C. `0.32cm`
D. `5xx10^-7m`

Explanation

To solve the problem, we need to find the separation between two equipotential surfaces given the surface charge density and the potential difference. Here's the step-by-step solution:

### Step 1: Understand the Electric Field due to an Infinite Sheet of Charge
The electric field (E) due to an infinite sheet of charge with surface charge density (σ) is given by the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where:
- \( \sigma = 10^{-7} \, C/m^2 \) (surface charge density)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \) (permittivity of free space)

### Step 2: Calculate the Electric Field
Substituting the value of σ into the formula:
\[ E = \frac{10^{-7}}{2 \times 8.85 \times 10^{-12}} \]
Calculating this gives:
\[ E = \frac{10^{-7}}{1.77 \times 10^{-11}} \approx 5.64 \times 10^{3} \, N/C \]

### Step 3: Relate Electric Field to Potential Difference and Distance
The relationship between electric field (E), potential difference (V), and distance (d) is given by:
\[ V = E \cdot d \]
In this case, we know the potential difference (V) is 5 V. Therefore, we can rearrange the equation to find the distance (d):
\[ d = \frac{V}{E} \]

### Step 4: Substitute the Values
Substituting the values of V and E into the equation:
\[ d = \frac{5 \, V}{5.64 \times 10^{3} \, N/C} \]

### Step 5: Calculate the Distance
Calculating this gives:
\[ d \approx \frac{5}{5.64 \times 10^{3}} \approx 8.86 \times 10^{-4} \, m \]
Converting this to millimeters:
\[ d \approx 0.886 \, mm \]

### Final Answer
The separation between the two equipotential surfaces is approximately:
\[ d \approx 0.886 \, mm \]