Let's work through this problem step-by-step to find the magnitude of the electric force on the charge [tex]\( q_1 \)[/tex]:
1. Determine the surface charge density ([tex]\( \sigma \)[/tex]):
The surface charge density is given by the charge on the plates divided by the area of the plates.
[tex]\[
\sigma = \frac{Q_{\text{plates}}}{A}
\][/tex]
Here, [tex]\( Q_{\text{plates}} = 7.07 \times 10^{-7} \, \text{C} \)[/tex] and [tex]\( A = 2.34 \times 10^{-3} \, \text{m}^2 \)[/tex].
So,
[tex]\[
\sigma = \frac{7.07 \times 10^{-7} \, \text{C}}{2.34 \times 10^{-3} \, \text{m}^2} \approx 0.0003021 \, \text{C/m}^2
\][/tex]
2. Calculate the electric field ([tex]\( E \)[/tex]):
The electric field between the plates can be found using the relation:
[tex]\[
E = \frac{\sigma}{\epsilon_0}
\][/tex]
where [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space, [tex]\( \epsilon_0 = 8.854187817 \times 10^{-12} \, \text{F/m} \)[/tex].
Substituting the values:
[tex]\[
E = \frac{0.0003021 \, \text{C/m}^2}{8.854187817 \times 10^{-12} \, \text{F/m}} \approx 34123598.7 \, \text{N/C}
\][/tex]
3. Calculate the electric force ([tex]\( F \)[/tex]) on [tex]\( q_1 \)[/tex]:
The force on a charge in an electric field is given by:
[tex]\[
F = q_1 \times E
\][/tex]
Here, [tex]\( q_1 = 6.62 \times 10^{-5} \, \text{C} \)[/tex] and [tex]\( E \approx 34123598.7 \, \text{N/C} \)[/tex].
So,
[tex]\[
F = 6.62 \times 10^{-5} \, \text{C} \times 34123598.7 \, \text{N/C} \approx 2258.98 \, \text{N}
\][/tex]
Thus, the magnitude of the electric force on [tex]\( q_1 \)[/tex] is:
[tex]\[
2258.98 \, \text{N}
\][/tex]
