Let's solve the problem of finding the electric field between two parallel plates with equal and opposite charges.
1. Given Values:
- Area of the plates, [tex]\( A = 5.68 \cdot 10^{-3} \, \text{m}^2 \)[/tex]
- Charge on the plates, [tex]\( Q = 4.38 \cdot 10^{-11} \, \text{C} \)[/tex]
- Permittivity of free space, [tex]\( \epsilon_0 = 8.854 \cdot 10^{-12} \, \text{N} \cdot \text{m}^2 \cdot \text{C}^{-2} \)[/tex]
2. Surface Charge Density ([tex]\( \sigma \)[/tex]):
The surface charge density is the charge per unit area:
[tex]\[
\sigma = \frac{Q}{A}
\][/tex]
Plugging the given values into the formula, we have:
[tex]\[
\sigma = \frac{4.38 \cdot 10^{-11} \, \text{C}}{5.68 \cdot 10^{-3} \, \text{m}^2}
\][/tex]
Evaluating this gives:
[tex]\[
\sigma = 7.711267605633803 \cdot 10^{-9} \, \text{C} \cdot \text{m}^{-2}
\][/tex]
3. Electric Field ([tex]\( E \)[/tex]):
The electric field between the plates can be determined using the relation:
[tex]\[
E = \frac{\sigma}{\epsilon_0}
\][/tex]
Substituting the values of [tex]\( \sigma \)[/tex] and [tex]\( \epsilon_0 \)[/tex]:
[tex]\[
E = \frac{7.711267605633803 \cdot 10^{-9} \, \text{C} \cdot \text{m}^{-2}}{8.854 \cdot 10^{-12} \, \text{N} \cdot \text{m}^2 \cdot \text{C}^{-2}}
\][/tex]
This simplifies to:
[tex]\[
E = 870.9360295497858 \, \text{N} / \text{C}
\][/tex]
Therefore, the electric field between the plates is approximately [tex]\( 870.936 \, \text{N/C} \)[/tex].
