Question

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 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 \(1.50 \mathrm{~V}\), 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 battery?

Short answer

In summary, for the given specifications: a) The area of each plate needed with a potential difference of 1.50 V is \(3.41 \times 10^8 m^2\). b) The area of each plate needed with the maximum potential difference of 3000 V is \(8.54 \times 10^{-5} m^2\). c) Neither capacitor is a practical replacement for a AAA battery, as the required plate area for (a) is too large, and the potential difference needed for (b) is too high for typical electronic devices.

Step-by-step solution

a) Finding the area of each plate with the potential difference of 1.50 V

To find the area of each plate, we'll first find the capacitance using the energy formula for capacitors: \(E_{stored} = \frac{1}{2} C V^2\) We are given the energy stored (3400 J) and potential difference (1.50 V). Rearrange the formula to find the capacitance: \(C = \frac{2E_{stored}}{V^2} = \frac{2(3400)}{(1.50)^2} = 3022.22 F\) Now we will find the area of each plate using the basic formula for the capacitance of a parallel plate capacitor: \(C = \frac{\epsilon_0 A}{d}\) where \(\epsilon_0 = 8.85 \times 10^{-12} F/m\) is the vacuum permittivity, A is the area of each plate, and d is the plate separation (0.001 m). Rearrange the formula to find the area: \(A = \frac{Cd}{\epsilon_0} = \frac{(3022.22)(0.001)}{(8.85 \times 10^{-12})} = 3.41 \times 10^8 m^2\)

b) Finding the area of each plate with the maximum potential difference

First, let's find the maximum potential difference across the capacitor using the dielectric breakdown voltage of air, which is \(3 \times 10^6 V/m\). Since the plate separation is 0.001 m, the maximum potential difference is: \(V_{max} = (3 \times 10^6)(0.001) = 3000 V\) Now we will find the capacitance, using the energy formula for capacitors: \(C = \frac{2E_{stored}}{V^2} = \frac{2(3400)}{(3000)^2} = 7.56 \times 10^{-4} F\) Now, we'll find the area of each plate using the same formula for the capacitance of a parallel plate capacitor: \(A = \frac{Cd}{\epsilon_0} = \frac{(7.56 \times 10^{-4})(0.001)}{(8.85 \times 10^{-12})} = 8.54 \times 10^{-5} m^2\)

c) Discussing practicality as a replacement for the AAA battery

The area of each plate needed for the capacitor in part a) is \(3.41 \times 10^8 m^2\). This area is much larger than the physical size of a AAA battery, making it impractical as a replacement. The area of each plate needed for the capacitor in part b) is \(8.54 \times 10^{-5} m^2\). Although the area needed has significantly reduced, the potential difference needed for this capacitor is 3000 V, which is quite large compared to typical electronic devices using a AAA battery (usually below 5 V). This makes it impractical as a replacement as well. Hence, neither capacitor is a practical replacement for the AAA battery.