Question

The energy stored in a fully charged capacitor is given by \(U=\frac{1}{2} C V^{2} .\) In a typical cardiac defibrillator, a capacitor charged to \(7500 \mathrm{V}\) has a stored energy of \(400 \mathrm{J}\). Based on this information, what is the charge on the capacitor in the cardiac defibrillator? a. \(1.1 \times 10^{-5} \mathrm{C}\) b. \(5 \times 10^{-2} \mathrm{C}\) c. \(1.1 \times 10^{-1} \mathrm{C}\) d. \(3.1 \times 10^{6} \mathrm{C}\)

Short answer

1.1 \times 10^{-1} \text{C}

Step-by-step solution

- Identify known variables

Identify the given values from the problem. The voltage across the capacitor is \(V = 7500 \, \text{V}\) and the energy stored in the capacitor is \(U = 400 \, \text{J}\).

- Use the energy formula

Use the formula for the energy stored in a capacitor, which is given by \( U = \frac{1}{2} C V^2 \). Rearrange the formula to solve for the capacitance \(C\).

- Solve for Capacitance

Rearrange the formula to solve for \(C\): \( C = \frac{2U}{V^2} \). Substituting the known values, we get \( C = \frac{2 \times 400 \, \text{J}}{(7500 \, \text{V})^2} \).

- Calculate Capacitance

Calculate the capacitance: \[ C = \frac{800 \text{J}}{56250000 \, \text{V}^2} = 1.422 \times 10^{-5} \, \text{F} \]

- Use the capacitance formula to find charge

The formula for the charge \(Q\) on a capacitor is given by \( Q = CV \). Substituting the values for \(C\) and \(V\), we get \( Q = (1.422 \times 10^{-5} \, \text{F})(7500 \, \text{V}) \).

- Calculate the charge

Calculate the charge: \[ Q = 1.0665 \times 10^{-1} \, \text{C} \] which can be approximated to \(1.1 \times 10^{-1} \, \text{C}\).

Key concepts

capacitor energy formula

A capacitor stores electrical energy in an electric field. The energy stored in a capacitor is given by the formula: \[ U = \frac{1}{2} C V^2 \] where \( U \) is the energy in Joules (J), \( C \) is the capacitance in Farads (F), and \( V \) is the voltage across the capacitor in Volts (V).
This means the energy stored depends on both the capacitance and the square of the voltage.
It’s crucial to understand this relationship because it helps you to calculate how much energy a capacitor can store at a given voltage.
In practical applications, like a cardiac defibrillator, knowing the energy storage is important to ensure enough power to deliver a life-saving shock.

charge on a capacitor

The charge \( Q \) on a capacitor is determined using the formula: \( Q = C V \) where \( Q \) is the charge in Coulombs (C), \( C \) is the capacitance in Farads (F), and \( V \) is the voltage in Volts (V).
This formula illustrates a direct relationship between charge, capacitance, and voltage.
Understanding this concept is key because the amount of charge a capacitor can hold varies depending on its capacitance and the applied voltage.
For instance, in the defibrillator example, we first calculated the capacitance (\(C = 1.422 \times 10^{-5} \) F) using energy and voltage values.
Thereafter, we utilized the capacitance and voltage to find the charge (\(Q = 1.1 \times 10^{-1} \) C).
This step-by-step process helps develop a clear understanding of how the charge on a capacitor is derived from basic electrical quantities.

capacitor voltage

Voltage across a capacitor is a critical parameter influencing both energy storage and charge. By definition, voltage \( V \) is the electrical potential difference between two points. In capacitors, this voltage creates an electric field that enables energy storage. The practical implication of voltage can be seen in devices like defibrillators.
The voltage must be sufficiently high to store the required energy (like the 7500 V in our example) to ensure the device can deliver enough power.
Calculating energy and charge easily demonstrates why voltage regulation is vital in capacitor-based systems.
It's important to recognize that even slight variations in voltage can significantly impact the energy (as energy depends on \( V^2 \)) and the charge stored in the capacitor.
The interconnectedness of voltage, energy, and charge in capacitors forms the foundation for understanding more complex electronic systems.