Answer:
Part A: 40.12×10⁻¹² F
Part B: 1.6048×10⁻⁸ C.
Part C: 3.2096×10⁻⁶ J
Explanation:
Part A
The formula of the capacitance of a capacitor is given as,
C = εrε₀A/d............... Equation 1
Where, C = capacitance of the capacitor, A = Area of the plate, d = distance of separation of the plates, εr = dielectric constant, ε₀ = permitivity of free space.
Given: A = 17 cm² = 0.0017 m², d = 0.150 cm = 0.0015 m, εr = 4.00, ε₀ = 8.85×10⁻¹² F/m
Substitute into equation 1
C = 4(0.0017)(8.85×10⁻¹²)/(0.0015)
C = 40.12×10⁻¹² F.
Part B
Using
Q = CV........................ Equation 2
Where Q = charge on either plates, V = Voltage.
Given: V = 400 V, C = 40.12×10⁻¹² F
Substitute into equation 3
Q = 400(40.12×10⁻¹²)
Q = 1.6048×10⁻⁸ C.
Part C
The formula for the energy stores in a charged capacitor is given as
U = 1/2CV²..................... Equation 3
Where U = Energy stored in the capacitor
Given: C = 40.12×10⁻¹² F, V = 400 V
Substitute into equation 3
U = 1/2(40.12×10⁻¹²)(400²)
U = 3.2096×10⁻⁶ J