a) Determine the capacitance, in pF, of a paper-filled parallel-plate capacitor (dielectric constant=3.70) having a plate area of 5.40cm2 and plate separation of 590 µm. Use 8.85 x 10-12 F/m for εo, the permittivity of free space.

For this I got 2998.53x10-2

b) Determine the maximum voltage, in kV, that can be applied to the capacitor. The dielectric strength of paper is 16.00 MV/m.

a) To determine the capacitance of a parallel-plate capacitor with a paper dielectric, we can use the formula:

C = (εo * εr * A) / d

where:
C = capacitance
εo = permittivity of free space = 8.85 x 10^(-12) F/m
εr = relative permittivity (dielectric constant) = 3.70
A = plate area = 5.40 cm^2 = 5.40 x 10^(-4) m^2
d = plate separation = 590 µm = 590 x 10^(-6) m

Substituting the values into the formula:

C = (8.85 x 10^(-12) F/m * 3.70 * 5.40 x 10^(-4) m^2) / (590 x 10^(-6) m)
C = 2998.53 x 10^(-12) F = 2998.53 pF

Therefore, the capacitance of the paper-filled parallel-plate capacitor is approximately 2998.53 pF.

b) To determine the maximum voltage that can be applied to the capacitor, we can use the formula:

V = εr * E_max

where:
V = maximum voltage
εr = relative permittivity (dielectric constant) = 3.70
E_max = maximum electric field strength = dielectric strength = 16.00 MV/m = 16.00 x 10^6 V/m

Substituting the values into the formula:

V = 3.70 * 16.00 x 10^6 V/m = 59.20 MV

Converting MV to kV:

V = 59.20 MV = 59.20 x 10^3 kV

Therefore, the maximum voltage that can be applied to the capacitor is approximately 59.20 kV.

a) To determine the capacitance of a paper-filled parallel-plate capacitor, you can use the formula:

C = (εo * εr * A) / d

where:
C is the capacitance
εo is the permittivity of free space (8.85 x 10^-12 F/m)
εr is the relative permittivity or dielectric constant of the paper (3.70)
A is the plate area (converted to square meters)
d is the plate separation (converted to meters)

First, convert the plate area from cm^2 to m^2:
1 cm^2 = 0.0001 m^2
plate area = 5.40 cm^2 * 0.0001 m^2/cm^2 = 0.00054 m^2

Next, convert the plate separation from µm to m:
1 µm = 0.000001 m
plate separation = 590 µm * 0.000001 m/µm = 0.00059 m

Now, substitute the values into the formula:

C = (8.85 x 10^-12 F/m * 3.70 * 0.00054 m^2) / 0.00059 m)
= 8.85 x 10^-12 F/m * 3.70 * 0.00054 m / 0.00059 m
≈ 8.13 x 10^-14 F

To convert from Farads (F) to picofarads (pF), you multiply by 10^12:

C ≈ 8.13 x 10^-14 F * 10^12 pF/F
≈ 81.3 pF

Therefore, the capacitance of the paper-filled parallel-plate capacitor is approximately 81.3 pF.

b) To determine the maximum voltage that can be applied to the capacitor, you can use the formula:

V_max = εr * V_breakdown

where:
V_max is the maximum voltage
εr is the relative permittivity or dielectric constant of the paper (3.70)
V_breakdown is the dielectric strength of the paper (16.00 MV/m)

Substitute the given values into the formula:

V_max = 3.70 * 16.00 MV/m

To convert from volts (V) to kilovolts (kV), divide by 10^6:

V_max = (3.70 * 16.00 MV/m) / (10^6 V/kV)
= 59.2 kV

Therefore, the maximum voltage that can be applied to the capacitor is 59.2 kV.