To find the spacing between the plates of a parallel plate capacitor given the potential difference and surface charge density, we can follow these steps:
### Step 1: Understand the given values
- Potential difference (V) = 100 V
- Surface charge density (σ) = 50 nC/cm²
### Step 2: Convert surface charge density to SI units
Surface charge density in SI units is expressed in coulombs per square meter (C/m²).
1 nC = \( 10^{-9} \) C, and \( 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 \).
Thus,
\[
\sigma = 50 \, \text{nC/cm}^2 = 50 \times 10^{-9} \, \text{C} / (10^{-4} \, \text{m}^2) = 50 \times 10^{-5} \, \text{C/m}^2 = 5.0 \times 10^{-5} \, \text{C/m}^2
\]
### Step 3: Use the relationship between charge, capacitance, and potential difference
The capacitance (C) of a parallel plate capacitor is given by:
\[
C = \frac{\varepsilon_0 A}{d}
\]
where:
- \( \varepsilon_0 \) = permittivity of free space \( = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- A = area of the plates
- d = distance between the plates
The charge (Q) on the plates can also be expressed as:
\[
Q = \sigma A
\]
### Step 4: Relate charge to capacitance and potential difference
Using the relationship \( Q = C \cdot V \):
\[
\sigma A = \frac{\varepsilon_0 A}{d} \cdot V
\]
### Step 5: Cancel A and rearrange to find d
By canceling A from both sides, we have:
\[
\sigma = \frac{\varepsilon_0 V}{d}
\]
Rearranging gives:
\[
d = \frac{\varepsilon_0 V}{\sigma}
\]
### Step 6: Substitute the known values
Substituting the known values into the equation:
\[
d = \frac{(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2) \cdot (100 \, \text{V})}{5.0 \times 10^{-5} \, \text{C/m}^2}
\]
### Step 7: Calculate d
Calculating the above expression:
\[
d = \frac{8.85 \times 10^{-10}}{5.0 \times 10^{-5}} = 1.77 \times 10^{-5} \, \text{m} = 177 \, \text{µm} = 177 \, \text{nm}
\]
### Final Answer
The spacing between the plates is \( 177 \, \text{nm} \).
---
