Answer :
To calculate the capacitance of a parallel-plate capacitor and the charge on its plates when a potential difference is applied, we'll follow these steps:
(a) Calculate the capacitance.
The capacitance [tex]C[/tex] of a parallel-plate capacitor is given by the formula:
[tex]C = \frac{\varepsilon_0 \cdot A}{d}[/tex]
Where:
- [tex]\varepsilon_0[/tex] is the vacuum permittivity, approximately [tex]8.85 \times 10^{-12} \ \text{F/m}[/tex].
- [tex]A[/tex] is the area of one of the plates.
- [tex]d[/tex] is the separation between the plates.
Calculate the area [tex]A[/tex]: The plates are circular, so the area [tex]A[/tex] is:
[tex]A = \pi r^2[/tex]
With [tex]r = 7.54 \ \text{cm} = 0.0754 \ \text{m}[/tex]:
[tex]A = \pi (0.0754)^2
\approx 0.0179 \ \text{m}^2[/tex]Plug the known values into the formula:
The separation [tex]d = 1.71 \ \text{mm} = 0.00171 \ \text{m}[/tex].
[tex]C = \frac{(8.85 \times 10^{-12} \ \text{F/m}) \times 0.0179 \ \text{m}^2}{0.00171 \ \text{m}}
\approx 9.25 \times 10^{-12} \ \text{F} \text{ or } 9.25 \ \text{pF}[/tex]
(b) What charge will appear on the plates if a potential difference of 97.7 V is applied?
The charge [tex]Q[/tex] on the plates is related to the capacitance and the potential difference [tex]V[/tex] by the formula:
[tex]Q = C \cdot V[/tex]
Using the capacitance we calculated:
[tex]Q = (9.25 \times 10^{-12} \ \text{F}) \times 97.7 \ \text{V}
\approx 9.03 \times 10^{-10} \ \text{C} \text{ or } 903 \ \text{pC}[/tex]
So, the capacitance of the capacitor is approximately [tex]9.25 \ \text{pF}[/tex], and the charge on the plates when a [tex]97.7 \ \text{V}[/tex] potential difference is applied is approximately [tex]903 \ \text{pC}[/tex].