Question

A parallel-plate capacitor is made from two plates 15.0 cm on each side and 4.65 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.25. An 19.5 V battery is connected across the plates.

1. What is the capacitance of this combination? C=____F

2. How much energy is stored in the capacitor? U=____J

3. If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor? U=___J

Answer 1

Area of the plates, A = 0.15 x 0.15 = 0.0225 m²

Separation between the plates, d = 4.65 mm = 0.00465 m

Potential applied to the capacitor, V = 19.5 V

1.

Half of the space between the plates contain air and this constitute an air capacitor of capacitance, C1 = ε₀ A/d/2 = 2 ε₀ A/d = 2 x 8.854 x 10^-12 x 0.0225/0.00465 = 85.68 x 10^-12 F

The other half is filed with Plexiglas of dielectric constant k = 3.25. this constitute a capacitor of capacitance,C2 =2 k ε₀ A/d = 2 x 3.25 x 8.854 x 10^-12 x 0.0225/0.00465 = 278.47 x 10^-12 F

The two parts of capacitors are in series and the total capacitance, C = C1C2/(C1 + C2)

C = 85.68 x 10^-12 x 278.47 x 10^-12 /(85.68 x 10^-12 +278.47 x 10^-12 ) = 65.52 x 10^-12 F

The capacitance of the combination, C = 65.52 x 10^-12 F

2.

Energy stored in the capacitor, U = ½ x CV² = ½ x 65.52 x 10^-12 x 19.5 x 19.5 = 1.24 x 10 ^-8 J

3.

If Plexiglas is removed the capacitance,

C = ε₀ A/d = 8.854 x 10^-12 x 0.0225/0.00465 = 42.84 x10^-12 F

Energy stored in the capacitor, U = ½ x CV² = ½ x 42.84 x 10^-12 x 19.5 x 19.5 = 0.81 x 10 ^-8 J