Certainly! Let's proceed step-by-step to find the electric field between the two parallel plates.
1. Identify the given values:
- Area of each plate, [tex]\( A = 5.68 \times 10^{-4} \, \text{m}^2 \)[/tex]
- Charge on the plates, [tex]\( Q = 8.38 \times 10^{-11} \, \text{C} \)[/tex]
- Permittivity of free space, [tex]\( \epsilon_0 = 8.854187817 \times 10^{-12} \, \text{F/m} \)[/tex] (Farads per meter)
2. Determine the surface charge density:
[tex]\[
\sigma = \frac{Q}{A}
\][/tex]
where [tex]\(\sigma\)[/tex] is the surface charge density (charge per unit area).
Plugging in the values:
[tex]\[
\sigma = \frac{8.38 \times 10^{-11} \, \text{C}}{5.68 \times 10^{-4} \, \text{m}^2}
= 1.475352112676056 \times 10^{-7} \, \text{C/m}^2
\][/tex]
3. Calculate the electric field [tex]\( E \)[/tex]:
The electric field [tex]\( E \)[/tex] between the plates is given by the formula:
[tex]\[
E = \frac{\sigma}{\epsilon_0}
\][/tex]
Substituting the values:
[tex]\[
E = \frac{1.475352112676056 \times 10^{-7} \, \text{C/m}^2}{8.854187817 \times 10^{-12} \, \text{F/m}}
\][/tex]
[tex]\[
E = 16662.76052833877 \, \text{N/C}
\][/tex]
Therefore, the electric field between the plates is [tex]\( 16662.76052833877 \, \text{N/C} \)[/tex].
