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 \(1.5 \mathrm{~V}\).) 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?

Short answer

Answer: The area of each plate is smaller in case b) because we are considering a higher voltage (3 kV) to avoid dielectric breakdown. However, neither capacitor is a practical replacement for a AAA battery because the required areas of the plates are still too large, and the voltage used in case b is very high and impractical for general applications.

Step-by-step solution

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: \(C = \dfrac{\epsilon_0 A}{d}\)

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: \(A = \dfrac{C d}{\epsilon_0} \approx \dfrac{(3022.22)(0.001)}{8.85\times10^{-12}} \approx 341070571.4 \mathrm{~m}^2\)

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: \(A = \dfrac{C d}{\epsilon_0} \approx \dfrac{(0.000756)(0.001)}{8.85 \times 10^{-12}} \approx 85.26 \mathrm{~m}^2\)

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.

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor is one of the simplest forms of capacitors, consisting of two conducting plates separated by a small distance. The space between the plates is usually filled with an insulating material or dielectric, like air in this case.

Capacitance, a key feature of a capacitor, measures its ability to store charge and is defined by the formula:

• \( C = \dfrac{\epsilon_0 A}{d} \)

where:

• \( C \) is the capacitance,
• \( \epsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \mathrm{~F/m}\)),
• \( A \) is the area of each plate,
• \( d \) is the separation between the plates.

With these plates, an electric field is created when a voltage is applied, and energy is stored in the field. The large area obtained in calculation, like the extensive \(341070571.4 \mathrm{~m}^2\) for case a, shows the impracticality due to physical size and space requirements for a typical application like a AAA battery.

Dielectric Breakdown

Dielectric breakdown is a phenomenon where an insulating material, like air, fails and becomes conductive when exposed to a high electric field. This happens when the field strength surpasses the material's threshold, allowing current to pass through where there should be none.

For air, this threshold is approximately 3 kV per mm. Hence, in the exercise, with a potential difference of 3 kV applied over 1 mm, dielectric breakdown is right on the edge of occurring.

• High voltages can cause air molecules to ionize, leading to sparks or arcing.
• The integrity of the capacitor is compromised when dielectric breakdown occurs.
• Sustaining such voltages make capacitors unsuitable for everyday battery replacements, since they require complex safety measures to prevent breakdown.

Understanding dielectric breakdown illustrates why smaller voltages (like 1.5V) are used in practical capacitor applications to ensure safety and reliability.

Energy Storage in Capacitors

The concept of energy storage in capacitors revolves around the ability to store electrical energy in the electric field between the plates. This stored energy (\(U\)) is described by:

• \(U = \dfrac{1}{2}CV^2\)

where \(C\) is the capacitance and \(V\) is the voltage applied across the plates.

In practical scenarios, the energy stored by capacitors is much smaller than batteries, as shown in the exercise where even large plate areas result in impractical capacitors for AAA battery replacement. Capacitors are typically utilized for:

• Sudden energy discharge requirements like starting motors or flashes.
• Smoothing out voltage in electronic circuits.

While they can rapidly discharge and recharge, their energy density is quite lower compared to chemical batteries, making them non-ideal for long-duration energy storage needs.