Question:

A parallel-plate capacitor has a plate area of 0.02 m² and a separation distance between the plates of 0.005 m. The space between the plates is filled with a dielectric material with a dielectric constant of 4. Determine the capacitance of the capacitor.

Answer:

The capacitance of a parallel-plate capacitor is given by the formula:

where:

• $C$ is the capacitance of the capacitor
• $\varepsilon_0$ is the permittivity of free space, approximately $8.85 \times 10^{-12} \, \text{F/m}$
• $\varepsilon_r$ is the relative permittivity (dielectric constant) of the material between the plates
• $A$ is the area of each plate
• $d$ is the separation distance between the plates

Plugging in the given values:

$A = 0.02 \, \text{m²}$ \ $d = 0.005 \, \text{m}$ \ $\varepsilon_r = 4$

The permittivity of free space $\varepsilon_0$ is a fundamental constant and does not change.

Substituting these values into the formula:

Calculating the expression:

Therefore, the capacitance of the parallel-plate capacitor is 7.08 microfarads.