Answer :
To determine the magnitude of the electric force on charge [tex]\( q_1 \)[/tex], we need to follow these steps:
1. Determine the Electric Field Between the Plates:
The electric field [tex]\( E \)[/tex] between two parallel plates is given by the formula:
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]
where:
- [tex]\( \sigma \)[/tex] is the surface charge density on the plates.
- [tex]\( \epsilon_0 \)[/tex] is the vacuum permittivity, which is approximately [tex]\( 8.854187817 \times 10^{-12} \, \text{F/m} \)[/tex].
The surface charge density [tex]\( \sigma \)[/tex] can be calculated using the charge [tex]\( Q \)[/tex] on the plates and the area [tex]\( A \)[/tex] of the plates:
[tex]\[ \sigma = \frac{Q}{A} \][/tex]
Given:
- [tex]\( Q = 5.83 \times 10^{-8} \, \text{C} \)[/tex]
- [tex]\( A = 7.34 \times 10^{-4} \, \text{m}^2 \)[/tex]
Plugging in these values:
[tex]\[ \sigma = \frac{5.83 \times 10^{-8} \, \text{C}}{7.34 \times 10^{-4} \, \text{m}^2} \approx 7.94 \times 10^{-5} \, \text{C/m}^2 \][/tex]
Now, the electric field [tex]\( E \)[/tex] is:
[tex]\[ E = \frac{\sigma}{\epsilon_0} = \frac{7.94 \times 10^{-5} \, \text{C/m}^2}{8.854187817 \times 10^{-12} \, \text{F/m}} \approx 8970646.95 \, \text{N/C} \][/tex]
2. Calculate the Force on the Charge [tex]\( q_1 \)[/tex]:
The force [tex]\( F \)[/tex] on a charge [tex]\( q_1 \)[/tex] in an electric field [tex]\( E \)[/tex] is given by:
[tex]\[ F = q_1 \times E \][/tex]
Given [tex]\( q_1 = 6.62 \times 10^{-6} \, \text{C} \)[/tex] and [tex]\( E \approx 8970646.95 \, \text{N/C} \)[/tex]:
[tex]\[ F = 6.62 \times 10^{-6} \, \text{C} \times 8970646.95 \, \text{N/C} \approx 59.39 \, \text{N} \][/tex]
Therefore, the magnitude of the electric force on [tex]\(q_1\)[/tex] is approximately [tex]\( 59.39 \, \text{N} \)[/tex].
1. Determine the Electric Field Between the Plates:
The electric field [tex]\( E \)[/tex] between two parallel plates is given by the formula:
[tex]\[ E = \frac{\sigma}{\epsilon_0} \][/tex]
where:
- [tex]\( \sigma \)[/tex] is the surface charge density on the plates.
- [tex]\( \epsilon_0 \)[/tex] is the vacuum permittivity, which is approximately [tex]\( 8.854187817 \times 10^{-12} \, \text{F/m} \)[/tex].
The surface charge density [tex]\( \sigma \)[/tex] can be calculated using the charge [tex]\( Q \)[/tex] on the plates and the area [tex]\( A \)[/tex] of the plates:
[tex]\[ \sigma = \frac{Q}{A} \][/tex]
Given:
- [tex]\( Q = 5.83 \times 10^{-8} \, \text{C} \)[/tex]
- [tex]\( A = 7.34 \times 10^{-4} \, \text{m}^2 \)[/tex]
Plugging in these values:
[tex]\[ \sigma = \frac{5.83 \times 10^{-8} \, \text{C}}{7.34 \times 10^{-4} \, \text{m}^2} \approx 7.94 \times 10^{-5} \, \text{C/m}^2 \][/tex]
Now, the electric field [tex]\( E \)[/tex] is:
[tex]\[ E = \frac{\sigma}{\epsilon_0} = \frac{7.94 \times 10^{-5} \, \text{C/m}^2}{8.854187817 \times 10^{-12} \, \text{F/m}} \approx 8970646.95 \, \text{N/C} \][/tex]
2. Calculate the Force on the Charge [tex]\( q_1 \)[/tex]:
The force [tex]\( F \)[/tex] on a charge [tex]\( q_1 \)[/tex] in an electric field [tex]\( E \)[/tex] is given by:
[tex]\[ F = q_1 \times E \][/tex]
Given [tex]\( q_1 = 6.62 \times 10^{-6} \, \text{C} \)[/tex] and [tex]\( E \approx 8970646.95 \, \text{N/C} \)[/tex]:
[tex]\[ F = 6.62 \times 10^{-6} \, \text{C} \times 8970646.95 \, \text{N/C} \approx 59.39 \, \text{N} \][/tex]
Therefore, the magnitude of the electric force on [tex]\(q_1\)[/tex] is approximately [tex]\( 59.39 \, \text{N} \)[/tex].
