Question

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?

Answer 1

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?

Answer:

(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

$A = \frac{C_oV}{\epsilon_o E}$

The energy density of stored energy is also given as;

U = ¹/₂KC₀V², U = 0.250 mJ

The charge stored in the capacitor;

$C_o =\frac{2U}{KV^2} \\\\C_o =\frac{2*0.25*10^{-3}}{2.6*(500)^2} = 7.692 *10^{-10} = 0.7692 \ nF$

Substitute these values and solve for Area

$A = \frac{C_oV}{\epsilon_o E} = \frac{7.692*10^{-10} *500}{8.85*10^{-12}*1.58*10^7} = 2.751*10^{-3} \ m^2$

A = 27.51 cm²