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

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

Step by step solution

01

Convert voltage to volts

Given voltage is 7.5 kV, which is equal to 7.5 * 1000 volts. So, \(V = 7500 V\)
02

Write down the energy stored in the capacitor

From the given information, the energy stored in the capacitor is \(W = 2400 J\).
03

Write down the formula for energy stored in a capacitor

The formula for the energy stored in a capacitor is given by \(W = \frac{1}{2}CV^2\).
04

Rearrange the formula to solve for capacitance

To find the capacitance, we need to rearrange the formula as follows: \(C = \frac{2W}{V^2}\)
05

Substitute the given values and calculate the capacitance

Plug in the values for energy and voltage into the formula and solve for the capacitance: \(C = \frac{2(2400)}{7500^2}\) \(C = \frac{4800}{56250000}\) \(C = 8.53333 \times 10^{-5} F\)
06

Write down the final answer

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel plate capacitor consisting of a pair of rectangular plates, each measuring \(1.00 \mathrm{~cm}\) by \(10.0 \mathrm{~cm},\) with a separation between the plates of \(0.100 \mathrm{~mm},\) is charged by a power supply at a potential difference of \(1.00 \cdot 10^{3} \mathrm{~V}\). The power supply is then removed, and without being discharged, the capacitor is placed in a vertical position over a container holding de-ionized water, with the short sides of the plates in contact with the water, as shown in the figure. Using energy considerations, show that the water will rise between the plates. Neglecting other effects, determine the system of equations that can be used to calculate the height to which the water rises between the plates. You do not have to solve the system.

A large parallel plate capacitor with plates that are square with side length \(1.00 \mathrm{~cm}\) and are separated by a distance of \(1.00 \mathrm{~mm}\) is dropped and damaged. Half of the areas of the two plates are pushed closer together to a distance of \(0.500 \mathrm{~mm}\). What is the capacitance of the damaged capacitor?

A \(4.00 \cdot 10^{3}-n F\) parallel plate capacitor is connected to a \(12.0-\mathrm{V}\) battery and charged. a) What is the charge \(Q\) on the positive plate of the capacitor? b) What is the electric potential energy stored in the capacitor? The \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor is then disconnected from the \(12.0-\mathrm{V}\) battery and used to charge three uncharged capacitors, a \(100 .-n F\) capacitor, a \(200 .-\mathrm{nF}\) capacitor, and a \(300 .-\mathrm{nF}\) capacitor, connected in series. c) After charging, what is the potential difference across each of the four capacitors? d) How much of the electrical energy stored in the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor was transferred to the other three capacitors?

You have an electric device containing a \(10.0-\mu \mathrm{F}\) capacitor, but an application requires an \(18.0-\mu \mathrm{F}\) capacitor. What modification can you make to your device to increase its capacitance to \(18.0-\mu \mathrm{F} ?\)

For a science project, a fourth. grader cuts the tops and bottoms off two soup cans of equal height, \(7.24 \mathrm{~cm},\) and with radii of \(3.02 \mathrm{~cm}\) and \(4.16 \mathrm{~cm},\) puts the smaller one inside the larger, and hot-glues them both on a sheet of plastic, as shown in the figure. Then she fills the gap between the cans with a special "soup' (dielectric constant of 63 ). What is the capacitance of this arrangement?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free