A parallel-plate capacitor has plates of area 0.20 m² separated by a distance of
1.0 mm. What is this capacitor's capacitance?

Question

A parallel-plate capacitor has plates of area 0.20 m² separated by a distance of 1.0 mm. What is this capacitor's capacitance?

Quizplus · Accepted Answer

The capacitance $C$ of a parallel-plate capacitor is given by the formula $C = \epsilon_0 \frac{A}{d}$, where $\epsilon_0$ is the vacuum permittivity ($8.85 \times 10^{-12} \, \text{F/m}$), $A$ is the area of the plates ($0.20 \, \text{m}^2$), and $d$ is the separation between the plates ($1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}$). Plugging in the values, $C = 8.85 \times 10^{-12} \, \text{F/m} \times \frac{0.20 \, \text{m}^2}{1.0 \times 10^{-3} \, \text{m}} = 1.77 \times 10^{-9} \, \text{F}$, which rounds to approximately $1.8 \, \text{nF}$.

A Parallel-Plate Capacitor Has Plates of Area 0 μ\muμ F

A Parallel-Plate Capacitor Has Plates of Area 0 $\mu$ F