A typical AAA battery has stored energy of about 3400 J. (Battery capacity is typically listed as \(625 \mathrm{~mA} \mathrm{~h}\), meaning that much charge can be delivered at approximately \(1.5 \mathrm{~V}\).) Suppose you want to build a parallel plate capacitor to store this amount of energy, using a plate separation of \(1.0 \mathrm{~mm}\) and with air filling the space between the plates. a) Assuming that the potential difference across the capacitor is \(1.5 \mathrm{~V},\) what must the area of each plate be? b) Assuming that the potential difference across the capacitor is the maximum that can be applied without dielectric breakdown occurring, what must the area of each plate be? c) Is either capacitor a practical replacement for the AAA batterv?

Step by step solution

01

Write the relevant formulas

We need formulas for the energy stored in a capacitor and the capacitance of a parallel plate capacitor. Energy stored in a capacitor: \(U = \dfrac{1}{2}CV^2\) Capacitance of a parallel plate capacitor: \(C = \dfrac{\epsilon_0 A}{d}\)

02

Solve for the area of each plate in case a

In this case, \(U = 3400 \mathrm{~J}\) and \(V = 1.5 \mathrm{~V}\) First, write the expression for the capacitance of the parallel plate capacitor using the energy formula: \(C = \dfrac{U}{\dfrac{1}{2}V^2} = \dfrac{2U}{V^2}\) Now, we can use this expression to get the capacitance: \(C = \dfrac{2(3400)}{1.5^2} \approx 3022.22 \mathrm{~F}\) Next, we can plug this value and the given separation (d = 0.001 m) in the capacitance formula to find the area (A) of each plate: \(A = \dfrac{C d}{\epsilon_0} \approx \dfrac{(3022.22)(0.001)}{8.85\times10^{-12}} \approx 341070571.4 \mathrm{~m}^2\)

03

Solve for the area of each plate in case b

In this case, \(U = 3400 \mathrm{~J}\) and \(V = 3 \mathrm{~kV}\) (dielectric breakdown voltage for air) Follow the same procedure as in Step 2 but with the new V value: \(C = \dfrac{2(3400)}{3000^2} \approx 0.000756 \mathrm{~F}\) Then plug this into the capacitance formula to find the area (A) of each plate: \(A = \dfrac{C d}{\epsilon_0} \approx \dfrac{(0.000756)(0.001)}{8.85 \times 10^{-12}} \approx 85.26 \mathrm{~m}^2\)

04

Comment on the practicality of replacing a AAA battery with either capacitor

In case a): The area required for each plate is very large (\(\approx 341070571.4 \mathrm{~m}^2\)). This makes it impractical to use the capacitor as a replacement for a AAA battery. In case b): The area required for each plate has significantly reduced compared to the previous case (\(\approx 85.26 \mathrm{~m}^2\)), but it is still too large to replace a AAA battery. Additionally, the required voltage of 3 kV for this case is very high and impractical for general applications. Therefore, neither capacitor is a practical replacement for a AAA battery.

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