A dielectric-filled parallel-plate capacitor has plate area A = 10.0cm2 , plate separation d = 7.00mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 7.50V . Throughout the problem, use ϵ0 = 8.85×10−12C2/N⋅m2 .
To solve this problem, we need to find the capacitance of the parallel-plate capacitor using the given information:
1. The plate area, A = 10.0 cm^2 = (10.0 × 10^(-4)) m^2.
2. The plate separation, d = 7.00 mm = (7.00 × 10^(-3)) m.
3. The dielectric constant, k = 3.00.
4. The value of the vacuum permittivity, ε₀ = 8.85 × 10^(-12) C^2/N·m^2.
The capacitance of a parallel-plate capacitor with a dielectric material can be calculated using the formula:
C = (k * ε₀ * A) / d
Now, let's substitute the values into the formula:
C = (3.00 * 8.85 × 10^(-12) C^2/N·m^2 * (10.0 × 10^(-4)) m^2) / (7.00 × 10^(-3)) m
C = (3.00 * 8.85 × 10^(-12) C^2/N·m^2 * 10.0 × 10^(-4)) / (7.00 × 10^(-3))
C = (3.00 * 8.85 × 10^(-12) * 10.0) / (7.00) N·m²/C
C = (3.00 * 8.85 × 10^(-11)) / (7.00) N·m²/C
C = 3.80 × 10^(-11) N·m²/C
Therefore, the capacitance of the dielectric-filled parallel-plate capacitor is 3.80 × 10^(-11) Farads.