Figure 17.31 b shows a thundercloud before a lightning strike has occurred. The bottom of the thundercloud and the Earth's surface might be modeled as a charged parallel plate capacitor. The base of the cloud, which is roughly parallel to the Earth's surface, serves as the negative plate and the region of Earth's surface under the cloud serves as the positive plate. The separation between the cloud base and the Earth's surface is small compared to the length of the cloud. (a) Find the capacitance for a thundercloud of base dimensions \(4.5 \mathrm{km}\) by \(2.5 \mathrm{km}\) located \(550 \mathrm{m}\) above the Earth's surface. (b) Find the energy stored in this capacitor if the charge magnitude is \(18 \mathrm{C}\).

To find the capacitance, we will use the formula for parallel plate capacitors: \(C = \frac{\epsilon_0 A}{d}\) Here, \(\epsilon_0\) is the vacuum permittivity (\(\epsilon_0 = 8.85 \times 10^{-12} F/m\)), \(A\) is the area of the base of the thundercloud, and \(d\) is the separation between the cloud base and the Earth's surface. From the problem, we have the base dimensions of the thundercloud as \(4.5 \mathrm{km}\) by \(2.5 \mathrm{km}\), and the separation as \(550 \mathrm{m}\). Firstly, we need to convert the dimensions into meters: \(4.5 \mathrm{km} = 4500 \mathrm{m}\) and \(2.5 \mathrm{km} = 2500\mathrm{m}\), then calculate the area \(A\): \(A = (4500 \mathrm{m}) \times (2500 \mathrm{m}) = 11,250,000 \mathrm{m}^2\) Now, apply the formula to find the capacitance: \(C = \frac{(8.85 \times 10^{-12} F/m)(11,250,000 \mathrm{m}^2)}{550 \mathrm{m}}\) \(C \approx 1.80 \times 10^{-7} F\)

Now, we will use the charge magnitude provided and the calculated capacitance to find the energy stored in the capacitor. The formula for the stored energy in a capacitor is: \(U = \frac{1}{2}CQ^2\) Here, \(U\) is the stored energy, \(C\) is the capacitance, and \(Q\) is the charge magnitude. Using the capacitance \(C \approx 1.80 \times 10^{-7} F\) and charge magnitude \(Q=18\mathrm{C}\): \(U = \frac{1}{2} (1.80 \times 10^{-7} F)(18 \mathrm{C})^2\) \(U \approx 29.16 \times 10^{3} J\) The capacitance of the thundercloud is approximately \(1.80 \times 10^{-7} F\), and the energy stored in the capacitor is approximately \(29.16 \times 10^{3} J\).