Sure, let's go through the problem step by step to determine the area of the plates.
Given:
- Distance between the plates, [tex]\( d = 8.08 \times 10^{-5} \)[/tex] meters
- Charge on the plates, [tex]\( Q = 2.24 \times 10^{-9} \)[/tex] coulombs
- Potential difference, [tex]\( V = 855 \)[/tex] volts
- Permittivity of free space, [tex]\( \varepsilon_0 = 8.85 \times 10^{-12} \)[/tex] farads per meter (F/m)
Firstly, we need to determine the capacitance [tex]\( C \)[/tex] using the formula for capacitance:
[tex]\[ C = \frac{Q}{V} \][/tex]
Plugging in the values,
[tex]\[ C = \frac{2.24 \times 10^{-9}}{855} \approx 2.6198830409356725 \times 10^{-12} \text{ farads} \][/tex]
Next, we use the formula for the capacitance of parallel plates to find the area [tex]\( A \)[/tex]:
[tex]\[ C = \frac{\varepsilon_0 \cdot A}{d} \][/tex]
Rearranging for the area [tex]\( A \)[/tex]:
[tex]\[ A = \frac{C \cdot d}{\varepsilon_0} \][/tex]
Plugging in the known values,
[tex]\[ A = \frac{2.6198830409356725 \times 10^{-12} \text{ F} \cdot 8.08 \times 10^{-5} \text{ m}}{8.85 \times 10^{-12} \text{ F/m}} \approx 2.3919384147751677 \times 10^{-5} \text{ m}^2 \][/tex]
Finally, we are asked to convert this area into units of [tex]\( 10^{-5} \text{ m}^2 \)[/tex]:
[tex]\[ \text{Area in units of } 10^{-5} \text{ m}^2 = 2.3919384147751677 \][/tex]
So, the area of the plates in units of [tex]\( 10^{-5} \text{ m}^2 \)[/tex] is approximately [tex]\( 2.3919384147751677 \)[/tex].
