Problem 51 - Physics

Question

The electric potential energy stored in the capacitor of a defibrillator is \(73 \mathrm{~J}\), and the capacitance is \(120 \mu \mathrm{F}\). What is the potential difference that exists across the capacitor plates?

Step-by-Step Solution

Answer

The potential difference is approximately 1103 volts.

Step 1: Understand the Relationship between Energy, Capacitance, and Voltage

The energy stored in a capacitor is given by the formula \( E = \frac{1}{2} C V^2 \), where \( E \) is the energy in joules, \( C \) is the capacitance in farads, and \( V \) is the potential difference in volts. We need to solve for \( V \).

Step 2: Rearrange the Formula to Solve for Voltage

We need to rearrange the formula \( E = \frac{1}{2} C V^2 \) to solve for \( V \). This gives us \( V = \sqrt{\frac{2E}{C}} \).

Step 3: Substitute the Given Values into the Formula

Substitute the given energy \( E = 73 \) J and capacitance \( C = 120 \mu F = 120 \times 10^{-6} \) F into the formula: \( V = \sqrt{\frac{2 \times 73}{120 \times 10^{-6}}} \).

Step 4: Calculate the Voltage

Perform the calculation: \( V = \sqrt{\frac{146}{120 \times 10^{-6}}} = \sqrt{\frac{146}{0.00012}} \). This simplifies to \( V = \sqrt{1216666.67} \), which approximately equals \( 1103 \) volts.

Key Concepts

Capacitors

Capacitors are electrical components that store and release energy. They do this by holding opposite charges on two conducting plates that are separated by an insulating material, known as the dielectric. When a voltage is applied across the two plates, one plate stores a positive charge, while the other stores a negative charge. The energy held is referred to as electric potential energy, which can be used when needed in electrical circuits. Capacitors are essential in various applications, like smoothing out power supplies, filtering signals, and in this specific example, they are used in medical equipment like defibrillators to provide the high-voltage pulse needed to reset a heart's rhythm.
Different types of capacitors exist depending on their size, range of capacitance, and dielectric material used. Each type offers unique characteristics suitable for various applications.

Voltage Calculation

Voltage calculation for a capacitor involves understanding the relationship between energy, capacitance, and the voltage across its plates. The key formula used here is \[ E = \frac{1}{2} C V^2 \]where \( E \) represents the energy stored in joules, \( C \) is the capacitance in farads, and \( V \) is the voltage in volts.
To find the voltage, you can rearrange this formula to solve for \( V \):\[ V = \sqrt{\frac{2E}{C}} \]
This calculation helps determine the potential difference across the capacitor plates, which is crucial for ensuring that systems like defibrillators have the required energy to function effectively. For example, if a specific energy level is needed, calculating the voltage allows engineers to design systems that meet those requirements.

Capacitance Formula

The capacitance \( C \) of a capacitor is a measure of its ability to store charge per unit voltage. It is mathematically expressed as the ratio of the charge \( Q \) on each plate to the voltage \( V \) across the plates:\[ C = \frac{Q}{V} \]
The unit of capacitance is the farad (F), named after Michael Faraday. The capacitance gives an idea of how much energy a capacitor can store — the higher the capacitance, the more charge at a given voltage.
In practical scenarios, capacitors range from picofarads (pF) in small circuits to farads (F) in large-scale applications. In exercises like the one we discussed, understanding the formula allows students to connect theoretical concepts with practical applications, helping them predict how a capacitor will behave in real-world scenarios based on its capacitance.