A capacitor is constructed from two aluminium plates 1.5cm on a side separated by 0.2mm of air.
(A) Calculate the capacitance of the capacitor. (2 sig figures)
(B)Calculate the voltage required to store 1nJ of energy in the capacitor.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html C=epsilon*area/separation
energy=1/2 C*V^2
To answer these questions, we need to use the formulas for capacitance and energy stored in a capacitor.
(A) Calculating the capacitance:
The capacitance of a parallel-plate capacitor is given by the formula:
C = (ε₀ * A) / d
Where:
- C is the capacitance
- ε₀ (epsilon naught) is the permittivity of free space (ε₀ ≈ 8.854 × 10⁻¹² F/m)
- A is the area of the plates
- d is the separation between the plates
In this case, the area of each plate (A) is (1.5 cm)² = 0.0225 m², and the separation between the plates (d) is 0.2 mm = 0.0002 m.
Plugging these values into the formula, we get:
C = (8.854 × 10⁻¹² F/m) * (0.0225 m²) / (0.0002 m)
C ≈ 9.8175 × 10⁻¹² F
Rounding to two significant figures, the capacitance of the capacitor is approximately 9.8 × 10⁻¹² F.
(B) Calculating the voltage required to store 1nJ of energy:
The energy stored in a capacitor is given by the formula:
E = (1/2) * C * V²
Where:
- E is the energy stored in the capacitor
- C is the capacitance
- V is the voltage across the capacitor
We are given the energy E = 1 nJ = 1 × 10⁻⁹ J, and we need to find the voltage V.
Rearranging the formula, we get:
V = √(2 * E / C)
Plugging in the values, we have:
V = √(2 * (1 × 10⁻⁹ J) / (9.8 × 10⁻¹² F))
V ≈ 0.02 V
Therefore, the voltage required to store 1 nJ of energy in the capacitor is approximately 0.02 V.