A parallel plate capacitor has square plates of side L= 12 cm and a distance d= 2 cm between the plates. Half of the capacitor areae is filled with a dielectric with k1 = 35. the other half is filled with a different dielectric with k2 = 8. The effective( equivalent capacitance if the capacitor is

Treat it as two capacitors in parallel, with areas of 72 cm^2 each, and dielectric constants of 8 and 35.

Add the two capacitances.

13.02

why 72cm^2?

To find the effective capacitance of the given capacitor, we need to calculate the capacitance of each half separately and then combine them.

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

C = (ε₀ * εr * A) / d

where C is the capacitance, ε₀ is the permittivity of free space (8.854 x 10^-12 F/m), εr is the relative permittivity (also known as the dielectric constant), A is the area of the plates, and d is the distance between the plates.

Let's calculate the capacitance for the first half of the capacitor:

The area of the first half (filled with dielectric k1) can be calculated as half of the total area of the capacitor:

A1 = (1/2) * (12 cm * 12 cm)

Now, we can calculate the capacitance using the given values:

C1 = (ε₀ * εr1 * A1) / d

where εr1 = k1 = 35 and d = 2 cm

Substituting the values, we get:

C1 = (8.854 x 10^-12 F/m * 35 * (1/2) * (0.12 m * 0.12 m)) / (0.02 m)

C1 ≈ 2.6 x 10^-11 F

Now, let's calculate the capacitance for the second half of the capacitor:

Using similar steps, we can calculate the area of the second half (filled with dielectric k2):

A2 = (1/2) * (12 cm * 12 cm)

Now, we can calculate the capacitance using the given values:

C2 = (ε₀ * εr2 * A2) / d

where εr2 = k2 = 8 and d = 2 cm

Substituting the values, we get:

C2 = (8.854 x 10^-12 F/m * 8 * (1/2) * (0.12 m * 0.12 m)) / (0.02 m)

C2 ≈ 9.4 x 10^-12 F

Finally, the effective capacitance (Ceq) of the capacitor is obtained by adding the capacitances of each half:

Ceq = C1 + C2

Ceq ≈ 2.6 x 10^-11 F + 9.4 x 10^-12 F

Ceq ≈ 3.5 x 10^-11 F

Therefore, the effective capacitance of the given capacitor is approximately 3.5 x 10^-11 Farads.