Plates of a parallel plate capacitor, having a potential difference 100 V applied across them, carry a surface charge density of `"50 nC cm"^(-2)`. Spacing between the plates is

Options

A. `329 mum`
B. `259mum`
C. `177mum`
D. `125mum`

Answer

To find the spacing between the plates of a parallel plate capacitor, we can use the relationship between the surface charge density (σ), the potential difference (V), and the permittivity of free space (ε₀).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Potential difference (V) = 100 V
- Surface charge density (σ) = 50 nC/cm² = 50 × 10⁻⁹ C/cm²

2. **Convert Surface Charge Density to SI Units:**
- Since 1 cm² = 10⁻⁴ m², we convert σ to C/m²:
\[ σ = 50 \times 10^{-9} \, \text{C/cm}^2 \times 10^4 \, \text{cm}^2/\text{m}^2 = 50 \times 10^{-5} \, \text{C/m}^2 = 5.0 \times 10^{-5} \, \text{C/m}^2 \]

3. **Use the Formula Relating Charge Density, Potential, and Plate Separation:**
- The electric field (E) between the plates of a parallel plate capacitor is given by:
\[ E = \frac{σ}{ε₀} \]
where ε₀ (permittivity of free space) = 8.85 × 10⁻¹² C²/(N·m²).

4. **Calculate the Electric Field (E):**
\[ E = \frac{5.0 \times 10^{-5}}{8.85 \times 10^{-12}} \approx 5.65 \times 10^{6} \, \text{N/C} \]

5. **Relate Electric Field to Potential Difference and Plate Separation:**
- The relationship between electric field (E), potential difference (V), and distance (d) is:
\[ E = \frac{V}{d} \]
Rearranging gives:
\[ d = \frac{V}{E} \]

6. **Substitute the Values to Find d:**
\[ d = \frac{100 \, \text{V}}{5.65 \times 10^{6} \, \text{N/C}} \approx 1.77 \times 10^{-5} \, \text{m} = 177 \, \mu m \]

### Final Answer:
The spacing between the plates of the parallel plate capacitor is approximately **177 micrometers (μm)**.