Question

The capacitor in an automatic external defibrillator is charged to \(7.5 \mathrm{kV}\) and stores \(2400 \mathrm{~J}\) of energy. What is its capacitance?

Short answer

Answer: The capacitance of the capacitor is approximately \(8.53 \times 10^{-5} F\).

Step-by-step solution

Convert voltage to volts

Given voltage is 7.5 kV, which is equal to 7.5 * 1000 volts. So, \(V = 7500 V\)

Write down the energy stored in the capacitor

From the given information, the energy stored in the capacitor is \(W = 2400 J\).

Write down the formula for energy stored in a capacitor

The formula for the energy stored in a capacitor is given by \(W = \frac{1}{2}CV^2\).

Rearrange the formula to solve for capacitance

To find the capacitance, we need to rearrange the formula as follows: \(C = \frac{2W}{V^2}\)

Substitute the given values and calculate the capacitance

Plug in the values for energy and voltage into the formula and solve for the capacitance: \(C = \frac{2(2400)}{7500^2}\) \(C = \frac{4800}{56250000}\) \(C = 8.53333 \times 10^{-5} F\)

Write down the final answer

The capacitance of the capacitor in the automatic external defibrillator is approximately \(8.53 \times 10^{-5} F\).

Key concepts

Electric Potential

The concept of electric potential is crucial when dealing with electric circuits and components like capacitors. Electric potential, often represented by the symbol \(V\), is a measure of the work needed to move a unit charge from a reference point to a specific point inside the field without producing any acceleration. It is usually measured in volts (V).

When discussing a capacitor, the electric potential between its plates is responsible for storing electric energy. This potential difference creates an electric field, essential for the capacitor's operation.

Key points to remember about electric potential in the context of capacitors include:

• It drives the flow of electric charges between the plates.
• The higher the potential, the more energy the capacitor can store.
• In defibrillators, a specific high potential allows for rapid discharge, crucial for medical applications.

Understanding electric potential provides insight into how efficiently a capacitor can store and release energy.

Energy Stored in Capacitor

Capacitors play a key role in storing electrical energy. The amount of energy a capacitor stores is related to its electric potential and capacitance. The energy stored in a capacitor can be calculated using the formula: \[W = \frac{1}{2}CV^2\]Where:

• \(W\) is the stored energy in joules (J).
• \(C\) is the capacitance in farads (F).
• \(V\) is the electric potential in volts (V).

This formula shows that the energy depends on the square of the voltage, highlighting why capacitors are effective for voltage stabilisation and filtering.

When applying this formula to a situation like in a defibrillator, knowing the energy and potential allows you to determine the necessary capacitance for effective operation. This is crucial for ensuring the device can deliver the required jolt to restart a heart. Thus, the concept of stored energy is foundational for designing capacitors in medical devices.

Capacitor Charge Formula

The charge on a capacitor is another fundamental aspect of its operation. The capacitor charge formula is given by:\[Q = C \times V\]Where:

• \(Q\) represents the charge in coulombs (C).
• \(C\) is the capacitance in farads (F).
• \(V\) is the potential difference in volts (V).

This equation illustrates the direct relationship between voltage and charge; as either the voltage or capacitance increases, so does the charge stored.

In the context of an automatic external defibrillator, knowing how much charge is stored helps ensure that the device is ready for use. This knowledge aids in designing capacitors that are reliable while delivering sufficient energy for lifesaving shocks. The capacitor charge formula thus serves as a key tool in guiding both capacitive design and application.