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Chapter 17: Problem 84

A defibrillator is used to restart a person's heart after it stops beating. Energy is delivered to the heart by discharging a capacitor through the body tissues near the heart. If the capacitance of the defibrillator is $9 \mu \mathrm{F}\( and the energy delivered is to be \)300 \mathrm{J},$ to what potential difference must the capacitor be charged?

Short Answer

Expert verified
Based on the given capacitance of a defibrillator and the energy it should deliver, calculate the potential difference that the capacitor must be charged to. Given: - Capacitance, C = 9 μF - Energy, E = 300 J Formula for energy stored in a capacitor: E = (1/2)CV^2 Solution: 1. Rearrange the formula to solve for potential difference, V: V = √(2E/C) 2. Substitute the given values for E and C: V = √(2 × 300J / (9 × 10^-6 F)) 3. Calculate the potential difference: V ≈ 2294.7 V The capacitor must be charged to a potential difference of approximately 2294.7 V to deliver the required energy.

Step by step solution

01

Write down the given information

We are given the following information: - Capacitance, C = \(9\mu F = 9 \times 10^{-6} F\) - Energy, E = \(300J\) The task is to find the potential difference, V.
02

Write down the formula for energy of a capacitor

The formula for the energy stored in a capacitor is: \(E = \frac{1}{2}CV^2\)
03

Solve for the potential difference

We can rearrange the formula to solve for the potential difference, V: \(V = \sqrt{\frac{2E}{C}}\)
04

Substitute the given values and calculate V

Now, we can plug the given values for E and C into the formula: \(V = \sqrt{\frac{2 \times 300J}{9 \times 10^{-6}F}}\)
05

Calculate the potential difference

After calculating the square root, we get: \(V \approx 2294.7 V\) So, the capacitor must be charged to a potential difference of approximately \(2294.7 V\) to deliver the required energy.

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Most popular questions from this chapter

An alpha particle (charge \(+2 e\) ) moves through a potential difference \(\Delta V=-0.50 \mathrm{kV} .\) Its initial kinetic energy is $1.20 \times 10^{-16} \mathrm{J} .$ What is its final kinetic energy?
In capacitive electrostimulation, electrodes are placed on opposite sides of a limb. A potential difference is applied to the electrodes, which is believed to be beneficial in treating bone defects and breaks. If the capacitance is measured to be \(0.59 \mathrm{pF},\) the electrodes are \(4.0 \mathrm{cm}^{2}\) in area, and the limb is \(3.0 \mathrm{cm}\) in diameter, what is the (average) dielectric constant of the tissue in the limb?
Capacitors are used in many applications where you need to supply a short burst of energy. A \(100.0-\mu \mathrm{F}\) capacitor in an electronic flash lamp supplies an average power of \(10.0 \mathrm{kW}\) to the lamp for $2.0 \mathrm{ms}$. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge?
A point charge \(q=-2.5 \mathrm{nC}\) is initially at rest adjacent to the negative plate of a capacitor. The charge per unit area on the plates is $4.0 \mu \mathrm{C} / \mathrm{m}^{2}\( and the space between the plates is \)6.0 \mathrm{mm} .$ (a) What is the potential difference between the plates? (b) What is the kinetic energy of the point charge just before it hits the positive plate, assuming no other forces act on it?
It has only been fairly recently that 1.0 -F capacitors have been readily available. A typical 1.0 -F capacitor can withstand up to \(5.00 \mathrm{V}\). To get an idea why it isn't easy to make a 1.0 -F capacitor, imagine making a \(1.0-\mathrm{F}\) parallel plate capacitor using titanium dioxide \((\kappa=90.0\) breakdown strength \(4.00 \mathrm{kV} / \mathrm{mm}\) ) as the dielectric. (a) Find the minimum thickness of the titanium dioxide such that the capacitor can withstand \(5.00 \mathrm{V}\). (b) Find the area of the plates so that the capacitance is \(1.0 \mathrm{F}\)
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