Very large capacitors have been considered as a means for storing electrical energy.If we constructed a very large parallel-plate capacitor of plate area 1.85 m² using paper (κ = 3.7)of thickness 1.5 mm as a dielectric,how much electrical energy would it store at a plate voltage of 1 500 V? (ε₀ = 8.85 × 10−¹² C²/N⋅m²) A)0.045 J B)9.1 J C)910 J D)1 800 J E)3 600 J

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Quizplus · Accepted Answer

The energy stored in a capacitor can be calculated using the formula $U = \frac{1}{2}CV^2$, where $U$ is the energy in joules, $C$ is the capacitance in farads, and $V$ is the voltage across the capacitor in volts. The capacitance $C$ of a parallel-plate capacitor is given by $C = \kappa\epsilon_0\frac{A}{d}$, where $\kappa$ is the dielectric constant of the material between the plates, $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2$), $A$ is the area of the plates in square meters, and $d$ is the separation between the plates in meters. Substituting the given values:- $\kappa = 3.7$- $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2$- $A = 1.85 \, \text{m}^2$- $d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}$- $V = 1500 \, \text{V}$First, calculate the capacitance $C$:$$C = 3.7 \times 8.85 \times 10^{-12} \frac{1.85}{1.5 \times 10^{-3}} = 3.7 \times 8.85 \times 10^{-12} \times \frac{1.85}{1.5 \times 10^{-3}} \approx 4.07 \times 10^{-8} \, \text{F}$$Then, calculate the energy $U$:$$U = \frac{1}{2} \times 4.07 \times 10^{-8} \times (1500)^2 \approx 0.045 \, \text{J}$$Therefore, the correct answer is 0.045 J.

Very Large Capacitors Have Been Considered as a Means for Storing