The SI unit of capacitance is the farad (F), named after the English physicist Michael Faraday.

The capacitance of a parallel-plate capacitor is directly proportional to the area of the plates, inversely proportional to the distance between the plates, and directly proportional to the permittivity of the material between the plates.

When a dielectric material is inserted between the plates of a capacitor, the capacitance increases because the dielectric material reduces the electric field between the plates.

The permittivity of a material is a measure of its ability to polarize in the presence of an electric field.

Ceramic materials have the highest permittivity among the given options.

The function of a dielectric material in a capacitor is to increase the capacitance by reducing the electric field between the plates.

The charge stored on a capacitor is given by the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Substituting the given values, we get Q = 10 μF * 12 V = 120 μC.

The energy stored in a capacitor is given by the equation E = 1/2 CV^2, where E is the energy, C is the capacitance, and V is the voltage. Substituting the given values, we get E = 1/2 * 20 μF * (10 V)^2 = 100 μJ.

The capacitance of a capacitor is inversely proportional to the frequency of an AC circuit.

The capacitance of a capacitor varies with temperature, depending on the material used as the dielectric.

The shape of the plates does not affect the capacitance of a capacitor.

The capacitance of a parallel-plate capacitor is given by the equation C = εA/d, where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the distance between the plates. Substituting the given values, we get C = (8.85 x 10^-12 F/m) * (10 cm^2) / (1 mm) = 8.85 pF.

The energy lost by a capacitor is given by the equation E = 1/2 C(V1^2 - V2^2), where E is the energy, C is the capacitance, V1 is the initial voltage, and V2 is the final voltage. Substituting the given values, we get E = 1/2 * C * (10 V)^2 - (8 V)^2 = 20 μJ.

The purpose of using a dielectric material in a capacitor is to increase the capacitance by reducing the electric field between the plates.

Ceramic materials are commonly used as dielectrics in capacitors due to their high permittivity and ability to withstand high voltages.