Question

A parallel plate capacitor has a capacitance of 8.8μF when filled with a dielectric. The area of each plate is 1.8 m² and the separation between the plates is 0.77×10⁻⁵ m. What is the dielectric constant of the dielectric?

Answer 1

The dielectric constant of the dielectric material in a parallel plate capacitor is calculated using the given capacitance, area of the plates, and separation between the plates. Upon calculation, the dielectric constant is found to be approximately 3.95.

Step-by-step explanation:

To determine the dielectric constant of the dielectric material in a parallel plate capacitor, we can use the formula for the capacitance of a capacitor with a dielectric:

C = (ε0 εr A) / d

where C is the capacitance, ε0 is the permittivity of free space, εr is the dielectric constant of the material, A is the area of one plate, and d is the separation between the plates.

The capacitance of the capacitor is given as 8.8μF, the area A is 1.8 m², and the separation d is 0.77×10−5 m. The permittivity of free space ε0 is 8.85×10−12 F/m. Plugging these values into the equation gives:

8.8×10−6 F = (8.85×10−12 F/m × εr × 1.8 m²) / (0.77×10−5 m)

Solving for εr gives:

εr = (8.8×10−6 × 0.77×10−5) / (8.85×10−12 × 1.8)

Calculating the above expression, we find that the dielectric constant (εr) is approximately 3.95.