Question

What is the capacitance of two square parallel plates 6.9 cm on a side that are separated by 1.9 mm of paraffin? The dielectric constant of paraffin is 2.2. Express your answer using two significant figures and include the appropriate units.

Answer 1

The capacitance of the two square parallel plates separated by 1.9 mm of paraffin with a dielectric constant of 2.2 is approximately 11 pF.

Step-by-step explanation:

To find the capacitance of two square parallel plates with a dielectric material in between, you can use the formula:

Capacitance, C = ε0 εrA/d

where ε0 is the permittivity of free space (ε0 = 8.85 × 10-12 F/m), εr is the relative permittivity (dielectric constant), A is the area of one plate in square meters, and d is the separation between the plates in meters.

In this case, the side of the plates is 6.9 cm, which we convert to meters (6.9 cm = 0.069 m). Thus, the area A is (0.069 m)^2. The separation d is 1.9 mm, which is 1.9 × 10-3 m. The dielectric constant εr is given as 2.2.

Substituting the values:

C = (8.85 × 10-12 F/m) × 2.2 × (0.069 m)^2 / (1.9 × 10-3 m)

Calculate the area of the plate:

A = (0.069 m)^2 = 0.004761 m^2

Now plug this value into the equation:

C = (8.85 × 10-12 F/m) × 2.2 × 0.004761 m^2 / (1.9 × 10-3 m)

After calculating the above expression, we get:

C = 1.14 × 10-11 F or 11 pF

To maintain two significant figures, the capacitance is approximately 11 pF (picoFarads).