When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 79 % of the dielectric strength. What is the area of each plate if the capacitor stores 0.250 mJ of energy under these conditions?

Complete question:

Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 × 10⁷ V/m . A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates.

a- When the electric field between the plates is 79 % of the dielectric strength, what is the energy density of the stored energy?

b- When the capacitor is connected to a battery with voltage 500.0 V , the electric field between the plates is 79 % of the dielectric strength. What is the area of each plate if the capacitor stores 0.25 mJ of energy under these conditions?

(a) The energy density of the stored energy is 2872.1 J/m³

(b) Area of the plate is 27.51 cm²

Explanation:

Given;

dielectric constant, K = 2.6

dielectric strength = 2.0 × 10⁷ V/m

potential difference of the battery, V = 500 V

Part (a)

Electric field strength, E = 79 % of dielectric strength

E = 0.79 x  2.0 × 10⁷ V/m

E = 1.58  × 10⁷ V/m

The energy density of the stored energy

u = ¹/₂Kε₀E²

where;

u is the energy density

ε₀ is permittivity of free space = 8.85 x 10⁻¹² C²/N.m²

u = ¹/₂Kε₀E² = ¹/₂ x 2.6 x 8.85 x 10⁻¹² x (1.58  × 10⁷)²

u = 28.721 x 10² J/m³

u = 2872.1 J/m³

Part (b) Area of the plate if the capacitor stores 0.250 mJ of energy

The energy density of stored energy is also given as;

U = ¹/₂KC₀V², U = 0.250 mJ

The charge stored in the capacitor;

Substitute these values and solve for Area

A = 27.51 cm²

answer; the correct answer is technician a; ///100%good luck

this could be newton#'s law of cooling. the cooling curve is exponential