Final answer:
The capacitance of a parallel plate capacitor with given dimensions and a dielectric constant for neoprene rubber is calculated to be about 5.61 pF, and it will hold a charge of approximately 50.5 pC when a voltage of 9.00 V is applied.
Step-by-step explanation:
Capacitance of a Parallel Plate Capacitor
To calculate the capacitance of a parallel plate capacitor with a dielectric material, you can use the following formula:
C = κε0A/d
where C is the capacitance, κ is the dielectric constant of the material between the plates, ε0 is the vacuum permittivity (ε0 ≈ 8.85 × 10-12 C2/N·m2), A is the area of the plates, and d is the separation between the plates.
In this problem, the area (A) is 1.7 cm2, which we convert to square meters (m2) for consistency with SI units. The area A is thus 1.7 × 10-4 m2. The separation (d) is 0.018 mm, or 18 × 10-6 meters. Given the dielectric constant (κ) for neoprene rubber is 6.7, we can calculate the capacitance as follows:
C = (6.7)(8.85 × 10-12 F/m)(1.7 × 10-4 m2) / (18 × 10-6 m) ≈ 5.61 × 10-12 F or 5.61 pF
If a voltage of 9.00 V is applied to this capacitor, the charge it holds can be calculated using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. So the charge will be:
Q = (5.61 × 10-12 F) (9.00 V) ≈ 5.05 × 10-11 C
Thus, when a voltage of 9.00 V is applied, the capacitor will hold a charge of approximately 50.5 pC.