To determine the separation between the plates of a parallel plate capacitor, we follow these steps:
1. Identify the given values:
- Area of the plates, [tex]\( A = 2.22 \times 10^{-4} \, \text{m}^2 \)[/tex]
- Charge on the plates, [tex]\( Q = 5.24 \times 10^{-9} \, \text{C} \)[/tex]
- Potential difference, [tex]\( V = 240 \, \text{V} \)[/tex]
- Permittivity of free space, [tex]\( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \)[/tex]
2. Calculate the capacitance:
The relationship between capacitance [tex]\( C \)[/tex], charge [tex]\( Q \)[/tex], and potential difference [tex]\( V \)[/tex] is given by:
[tex]\[
C = \frac{Q}{V}
\][/tex]
Plugging in the given values:
[tex]\[
C = \frac{5.24 \times 10^{-9} \, \text{C}}{240 \, \text{V}}
\][/tex]
Calculate [tex]\( C \)[/tex]:
[tex]\[
C = 2.183333 \times 10^{-11} \, \text{F} \quad (\text{farads})
\][/tex]
3. Use the formula for the capacitance of a parallel plate capacitor:
The formula is:
[tex]\[
C = \frac{\epsilon_0 \cdot A}{d}
\][/tex]
Where:
- [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space
- [tex]\( A \)[/tex] is the area of the plates
- [tex]\( d \)[/tex] is the separation between the plates
Rearranging to solve for [tex]\( d \)[/tex]:
[tex]\[
d = \frac{\epsilon_0 \cdot A}{C}
\][/tex]
4. Calculate the separation distance [tex]\( d \)[/tex]:
Substitute the known values:
[tex]\[
d = \frac{8.85 \times 10^{-12} \, \text{F/m} \times 2.22 \times 10^{-4} \, \text{m}^2}{2.183333 \times 10^{-11} \, \text{F}}
\][/tex]
Perform the calculation:
[tex]\[
d \approx 8.998625954198475 \times 10^{-5} \, \text{m}
\][/tex]
5. Express the answer:
The separation between the plates is approximately:
[tex]\[
8.998625954198475 \times 10^{-5} \, \text{m}
\][/tex]
When writing this in scientific notation to the required precision (given the power of [tex]\(10^{-5}\)[/tex]):
The answer is [tex]\( 8.998625954198475 \)[/tex].
Therefore, the separation between the plates is 8.998625954198475 [tex]\(\times 10^{-5}\)[/tex] meters.
