Sure! Let's break down this problem step by step in detail:
1. Given Data:
- Area of the plates, [tex]\( A = 7.34 \times 10^{-4} \ \text{m}^2 \)[/tex]
- Charge on the plates, [tex]\( \sigma = 5.83 \times 10^{-8} \ \text{C} \)[/tex]
- Charge placed between the plates, [tex]\( q_1 = 6.62 \times 10^{-6} \ \text{C} \)[/tex]
- Vacuum permittivity, [tex]\( \epsilon_0 = 8.854 \times 10^{-12} \ \text{C}^2/(\text{N} \cdot \text{m}^2) \)[/tex]
2. Calculate the Electric Field [tex]\( E \)[/tex]:
- First, let's find the surface charge density [tex]\( \sigma \)[/tex]:
[tex]\[
\sigma = \frac{\text{Charge on the plates}}{\text{Area of the plates}} = \frac{5.83 \times 10^{-8} \ \text{C}}{7.34 \times 10^{-4} \ \text{m}^2}
\][/tex]
[tex]\[
\sigma = 7.94 \times 10^{-5} \ \text{C}/\text{m}^2
\][/tex]
- Now, the electric field [tex]\( E \)[/tex] between two large parallel plates separated by a vacuum is given by:
[tex]\[
E = \frac{\sigma}{\epsilon_0} = \frac{7.94 \times 10^{-5} \ \text{C}/\text{m}^2}{8.854 \times 10^{-12} \ \text{C}^2/(\text{N} \cdot \text{m}^2)}
\][/tex]
Simplifying this, we get:
[tex]\[
E \approx 8970837.239 \ \text{N/C}
\][/tex]
3. Calculate the Electric Force [tex]\( F \)[/tex]:
- The force [tex]\( F \)[/tex] on a charge [tex]\( q \)[/tex] in an electric field [tex]\( E \)[/tex] is given by:
[tex]\[
F = q \cdot E = 6.62 \times 10^{-6} \ \text{C} \times 8970837.239 \ \text{N/C}
\][/tex]
Simplifying this, we get:
[tex]\[
F \approx 59.39 \ \text{N}
\][/tex]
Thus, the magnitude of the electric force on [tex]\( q_1 \)[/tex] is approximately [tex]\( 59.39 \ \text{N} \)[/tex].
