Answer :
To find the potential difference [tex]\( V \)[/tex] between the plates, we can follow these steps:
1. Determine the surface charge density [tex]\(\sigma\)[/tex]:
The surface charge density [tex]\(\sigma\)[/tex] is defined as the charge per unit area:
[tex]\[ \sigma = \frac{\text{charge}}{\text{area}} \][/tex]
Given:
- Charge [tex]\(Q = 5.24 \times 10^{-8} \, \text{C}\)[/tex]
- Area [tex]\(A = 8.22 \times 10^{-4} \, \text{m}^2\)[/tex]
Substituting the values:
[tex]\[ \sigma = \frac{5.24 \times 10^{-8}}{8.22 \times 10^{-4}} \approx 6.3747 \times 10^{-5} \, \text{C/m}^2 \][/tex]
2. Calculate the electric field [tex]\( E \)[/tex]:
The electric field [tex]\( E \)[/tex] between two parallel plates is given by [tex]\(\frac{\sigma}{\epsilon_0}\)[/tex], where [tex]\(\epsilon_0\)[/tex] is the permittivity of free space (vacuum permittivity):
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]
Given:
- [tex]\(\epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}\)[/tex]
- [tex]\(\sigma = 6.3747 \times 10^{-5} \, \text{C/m}^2\)[/tex]
Substituting the values:
[tex]\[ E = \frac{6.3747 \times 10^{-5}}{8.854 \times 10^{-12}} \approx 7199792.03 \, \text{N/C} \][/tex]
3. Determine the potential difference [tex]\( V \)[/tex]:
The potential difference [tex]\( V \)[/tex] between the plates is given by [tex]\( V = E \times \text{separation} \)[/tex], where the separation is the distance between the plates:
[tex]\[ V = E \times d \][/tex]
Given:
- Electric field [tex]\( E = 7199792.03 \, \text{N/C} \)[/tex]
- Separation [tex]\( d = 2.42 \times 10^{-5} \, \text{m} \)[/tex]
Substituting the values:
[tex]\[ V = 7199792.03 \times 2.42 \times 10^{-5} \approx 174.235 \, \text{V} \][/tex]
Therefore, the potential difference [tex]\( V \)[/tex] between the plates is approximately [tex]\( 174.235 \, \text{V} \)[/tex].
1. Determine the surface charge density [tex]\(\sigma\)[/tex]:
The surface charge density [tex]\(\sigma\)[/tex] is defined as the charge per unit area:
[tex]\[ \sigma = \frac{\text{charge}}{\text{area}} \][/tex]
Given:
- Charge [tex]\(Q = 5.24 \times 10^{-8} \, \text{C}\)[/tex]
- Area [tex]\(A = 8.22 \times 10^{-4} \, \text{m}^2\)[/tex]
Substituting the values:
[tex]\[ \sigma = \frac{5.24 \times 10^{-8}}{8.22 \times 10^{-4}} \approx 6.3747 \times 10^{-5} \, \text{C/m}^2 \][/tex]
2. Calculate the electric field [tex]\( E \)[/tex]:
The electric field [tex]\( E \)[/tex] between two parallel plates is given by [tex]\(\frac{\sigma}{\epsilon_0}\)[/tex], where [tex]\(\epsilon_0\)[/tex] is the permittivity of free space (vacuum permittivity):
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]
Given:
- [tex]\(\epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}\)[/tex]
- [tex]\(\sigma = 6.3747 \times 10^{-5} \, \text{C/m}^2\)[/tex]
Substituting the values:
[tex]\[ E = \frac{6.3747 \times 10^{-5}}{8.854 \times 10^{-12}} \approx 7199792.03 \, \text{N/C} \][/tex]
3. Determine the potential difference [tex]\( V \)[/tex]:
The potential difference [tex]\( V \)[/tex] between the plates is given by [tex]\( V = E \times \text{separation} \)[/tex], where the separation is the distance between the plates:
[tex]\[ V = E \times d \][/tex]
Given:
- Electric field [tex]\( E = 7199792.03 \, \text{N/C} \)[/tex]
- Separation [tex]\( d = 2.42 \times 10^{-5} \, \text{m} \)[/tex]
Substituting the values:
[tex]\[ V = 7199792.03 \times 2.42 \times 10^{-5} \approx 174.235 \, \text{V} \][/tex]
Therefore, the potential difference [tex]\( V \)[/tex] between the plates is approximately [tex]\( 174.235 \, \text{V} \)[/tex].
