Sure, let's walk through the step-by-step solution to find the charge on the plates.
1. Given Values:
[tex]\[
\text{Area (A)} = 8.22 \times 10^{-4} \, \text{m}^2
\][/tex]
[tex]\[
\text{Separation (d)} = 2.42 \times 10^{-5} \, \text{m}
\][/tex]
[tex]\[
\text{Potential Difference (V)} = 25.0 \, \text{V}
\][/tex]
[tex]\[
\text{Permittivity of Free Space (\( \epsilon_0 \))} = 8.85 \times 10^{-12} \, \text{F/m}
\][/tex]
2. Calculate the Capacitance (C):
The formula for capacitance of a parallel plate capacitor is:
[tex]\[
C = \frac{\epsilon_0 \cdot A}{d}
\][/tex]
Plugging in the values:
[tex]\[
C = \frac{8.85 \times 10^{-12} \, \text{F/m} \times 8.22 \times 10^{-4} \, \text{m}^2}{2.42 \times 10^{-5} \, \text{m}}
\][/tex]
Solving for [tex]\(C\)[/tex]:
[tex]\[
C = 3.0060743801652887 \times 10^{-10} \, \text{F}
\][/tex]
3. Calculate the Charge (Q):
The formula for the charge on a capacitor is:
[tex]\[
Q = C \times V
\][/tex]
Plugging in the values:
[tex]\[
Q = 3.0060743801652887 \times 10^{-10} \, \text{F} \times 25.0 \, \text{V}
\][/tex]
Solving for [tex]\(Q\)[/tex]:
[tex]\[
Q = 7.515185950413222 \times 10^{-9} \, \text{C}
\][/tex]
4. Convert the Charge to the Given Unit:
The charge is already in the form of [tex]\( Q \times 10^{-9} \, \text{C} \)[/tex], so we simply report the coefficient.
Thus, the charge on the plates is [tex]\( \boxed{7.515185950413222} \times 10^{-9} \, \text{C} \)[/tex].
Therefore, the answer is:
[tex]\[
\boxed{7.515185950413222}
\][/tex]
