A 3.0-pFcapacitor consists of two large closely-spaced parallel plates that have surface charge densities of $\pm 1.0 \mathrm { nC } / \mathrm { mm } ^ { 2 }$. If the potential across the plates is $23.0 \mathrm { kV }$ with only air between them, find the surface area of each of the plates. $\left( \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 } \right.$ )

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The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$, where $A$ is the surface area of the plates, $d$ is the distance between the plates, and $\varepsilon_0$ is the permittivity of free space. However, to find the area without the distance $d$, we use the relationship between capacitance, charge ($Q$), and potential difference ($V$): $C = \frac{Q}{V}$. Given $C = 3.0 \times 10^{-12} \, \text{F}$ and $V = 23.0 \times 10^3 \, \text{V}$, we find $Q = CV = 3.0 \times 10^{-12} \, \text{F} \times 23.0 \times 10^3 \, \text{V} = 6.9 \times 10^{-8} \, \text{C}$. The surface charge density $\sigma = 1.0 \, \text{nC/mm}^2 = 1.0 \times 10^{-9} \, \text{C/mm}^2$, so the area $A = \frac{Q}{\sigma} = \frac{6.9 \times 10^{-8} \, \text{C}}{1.0 \times 10^{-9} \, \text{C/mm}^2} = 69 \, \text{mm}^2$.

A 30-PFcapacitor Consists of Two Large Closely-Spaced Parallel Plates That Have