A medical defibrillator stores \(950 \mathrm{J}\) in a \(100-\mu \mathrm{F}\) capacitor. (a) What is the voltage across the capacitor? (b) If the capacitor discharges \(300 \mathrm{J}\) of its stored energy in \(2.5 \mathrm{ms}\), what's the power delivered during this time?

Capacitors are fundamental components in electronics, known for their ability to store energy in an electric field. The energy stored in a capacitor can be calculated using the formula \( E = 0.5 \times C \times V^2 \), where \( E \) is the energy in joules (J), \( C \) is the capacitance in farads (F), and \( V \) is the voltage across the capacitor in volts (V).

In our medical defibrillator example, a capacitor with 950 joules of energy and a capacitance of 100 microfarads (\(100 \, \mu \mathrm{F}\)) is used. This high amount of energy stored in such a small device demonstrates the efficiency of capacitors in energy storage applications, particularly when high-energy discharge is needed in a short period, like in medical emergencies.

Understanding how the energy is related to the voltage and capacitance helps us design systems where precise energy discharge is crucial. The square relationship between voltage and stored energy indicates that increasing the voltage can significantly increase the energy stored in the capacitor.

Capacitor discharge refers to the release of stored energy from the capacitor into a connected circuit. It is a rapid process where the electrical energy is converted into another form, such as kinetic energy in motors, or in this case, to another type of electrical energy capable of providing therapeutic effects in medical devices.

During discharge, the voltage across the capacitor's terminals decreases, and the current flows until the stored energy is depleted or until the circuit is interrupted. The issue with the defibrillator capacitor discharging 300 joules in 2.5 milliseconds gives us a glimpse of how quick and powerful these discharges can be, which is critical in applications requiring a swift and robust energy transfer.

For students and engineers alike, understanding the discharge characteristics of capacitors is essential for effectively utilizing them in circuits, especially when timing and the amount of energy released are vital factors in the system's performance.

The power delivered by a capacitor during discharge is a measure of how fast the stored energy is being converted into another form. It can be calculated with the formula \( P = \Delta E / \Delta t \), where \( P \) is power in watts (W), \( \Delta E \) is the change in energy in joules (J), and \( \Delta t \) is the time interval in seconds (s).

In our medical defibrillator scenario, the calculated power delivery is essential for ensuring the device operates effectively during critical moments. Computing the power as 300 joules divided by 2.5 milliseconds underscores how a high amount of energy can be delivered in a very short time frame — a characteristic integral to defibrillators designed to provide a life-saving electric shock.

Comprehending how power is delivered by capacitors not only supports students in solving physics problems but also provides insights into the practical aspects of electrical energy management in various technological applications.