a parallel plate capacitor has square plates that have edge length equal to 100cm and are separated by 1 mm. It is connected to a battery and is charged to 12V. How much energy is stored in the capacitor?

3.187*10⁻⁷ J.

Explanation:

The expression for the energy stored in a capacitor is given as,

E = 1/2CV² ... Equation 1

Where E = Energy stored in the capacitor, C = Capacitance of the capacitor, V = Voltage.

But,

C = ε₀A/d ... Equation 2

Where A = Area of the plates, d = distance of separation of the plates, ε₀ = permitivity of free space

Substitute equation 2 into equation 1

E = 1/2 (ε₀A/d) V² ... Equation 3

Given: V = 12 V, d = 1 mm = 0.001 m

Note Edge length is the diagonal of a square

From Pythagoras,

L²+L² = d

Where d = diagonal = 100 cm = 1 m

2L² = 1

L² = 1/2

L = √ (1/2) m

A = area of a square = L², where L = Length of the square plates = √ (1/2)

A = [√ (1/2) ]² = 1/2 m², ε₀ = 8.854*10⁻¹² F/m

Substitute into equation 3

E = 1/2 (8.854*10⁻¹²*1/2/0.001) 12²

E = 36 (8854*10⁻¹²)

E = 318744*10⁻¹²

E = 318744*10⁻⁷ J.

Hence the amount of energy stored in the capacitor = 3.187*10⁻⁷ J.

the energy is stored in the capacitor is 0.32 μJ

Explanation:

Given;

distance of separation, d = 1 mm = 0.001 m

edge length of the square = 100 cm

potential difference across the plates, V = 12 v

let the side of the square = L

This edge length is also the diagonal of the square which makes a right angle with the side of the square.

Applying Pythagoras theorem;

L² + L² = 100²

2L² = 100²

L² = 100²/2

Note area of a square is L²

A = L² = 100²/2 = 5000 cm²

A (m²) = 5000 cm² x 1m² / (100 cm) ²

A = 5000 cm² x 1m²/10000 cm²

A = 0.5 m²

Energy stored in a parallel plate capacitor, E = ¹/₂CV²

C = ε₀A/d

where;

ε₀ is permittivity of free space = 8.85 x 10⁻¹² F/m

d is the distance of separation = 0.001 m

A is the area of the plate

C = ε₀A/d = (8.85 x 10⁻¹²) x0.5 / 0.001

C = 4425 x 10⁻¹² F

E = ¹/₂CV² = ¹/₂ x 4425 x 10⁻¹² x (12) ²

E = 318600 x 10⁻¹² = 0.32 μJ

Therefore, the energy is stored in the capacitor is 0.32 μJ