Answer:
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
Answer:
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.
