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

Short Answer

Expert verified
The potential difference across the capacitor plates is approximately 1103.47 V.

Step by step solution

01

Understanding the Given Values

We are given that the electric potential energy (U) stored in the capacitor is 73 Joules, and the capacitance (C) is 120 microfarads, which can be written as \(120 \times 10^{-6} \; \mathrm{F}\). We need to find the potential difference (V) across the capacitor plates.
02

Using the Formula for Electric Potential Energy

The formula for the electric potential energy stored in a capacitor is \( U = \frac{1}{2} C V^2 \). To find the potential difference \( V \), we need to rearrange this formula.
03

Rearranging the Formula to Solve for V

From the equation \( U = \frac{1}{2} C V^2 \), we can solve for \( V \) by rearranging it: \[ V = \sqrt{\frac{2U}{C}} \] where \( U \) is the electric potential energy and \( C \) is the capacitance.
04

Substituting the Given Values

Now we substitute the given values into the equation: \[ V = \sqrt{\frac{2 \times 73}{120 \times 10^{-6}}} \].
05

Calculating the Potential Difference

When we compute the above expression: \( V = \sqrt{\frac{146}{120 \times 10^{-6}}} = \sqrt{\frac{146}{0.00012}} \). Calculating this gives \( V \approx \sqrt{1216666.67} \approx 1103.47 \; \mathrm{V} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance
Capacitance is a fundamental concept in electronics which describes a component's ability to store an electric charge. It is typically denoted by the letter \( C \) and is measured in farads (F). Understanding capacitance is crucial in many fields of electrical engineering and physics.
Capacitance depends mainly on two factors:
  • The surface area of the conductive plates.
  • The distance between those plates.
Moreover, the dielectric material placed between the plates can affect the capacitance. A higher permittivity of the dielectric will increase the capacitance.
In practice, capacitors usually have capacitances ranging from picofarads (\(10^{-12}\mathrm{F}\)) to microfarads (\(10^{-6}\mathrm{F}\)). When learning about electrical circuits, it's often helpful to become familiar with using microfarads and other small units. It's important to remember that capacitance alone doesn't determine how much energy is stored. The potential difference also plays a significant role.
Potential Difference
The potential difference, often referred to as voltage, is the work needed to move a charge from one point to another against an electric field. For capacitors, this potential difference \( V \) across the plates determines how much energy is stored within the system.
The potential difference is measured in volts (V), and it directly relates to the amount of energy that a capacitor can store. The formula \( U = \frac{1}{2} C V^2 \) links potential energy \( U \), capacitance \( C \), and the potential difference \( V \). Therefore, larger potential differences allow more energy to be stored given the same capacitance.
Physically speaking, voltage is a measure of the electrical "pressure" that pushes electrons through a circuit. Understanding the concept of potential difference is vital for grasping how electronic devices power various components and convert stored potential energy into useful work.
Capacitor
A capacitor is an essential electrical component that stores energy in the form of an electric field. Within circuits, capacitors serve many purposes:
  • They store and release electrical energy quickly to smooth out fluctuations in power supply.
  • Capacitors can block direct current (DC) while allowing alternating current (AC) to pass, effectively filtering circuits.
Capacitors are made up of two conductive plates separated by a dielectric (insulating material). When a voltage is applied across the plates, an electric field forms, and the capacitor begins storing energy. This stored energy can be released to other components when needed, acting like a temporary battery.
The design and application of a capacitor, such as in a defibrillator, demonstrate its critical role in modern electronics; for example, delivering a rapid discharge of energy precisely when and where it is required. Understanding how capacitors work helps in creating more efficient electronic devices that are capable of handling varying electrical loads.

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Most popular questions from this chapter

An axon is the relatively long tail-like part of a neuron. or nerve cell. The outer surface of the axon membrane (dielectric constant \(=5,\) thickness \(=1 \times 10^{-8} \mathrm{m}\) ) is charged positively, and the inner portion is charged negatively. Thus, the membrane is a kind of capacitor. Assuming that the membrane acts like a parallel plate capacitor with a plate area of \(5 \times 10^{-6} \mathrm{m}^{2},\) what is its capacitance?

The membrane that surrounds a certain type of living cell has a surface area of \(5.0 \times 10^{-9} \mathrm{m}^{2}\) and a thickness of \(1.0 \times 10^{-8} \mathrm{m} .\) Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of \(5.0 .\) (a) The potential on the outer surface of the membrane is \(+60.0 \mathrm{mV}\) greater than that on the inside surface. How much charge resides on the outer surface? (b) If the charge in part (a) is due to positive ions (charge \(+e\) ), how many such ions are present on the outer surface?

The potential difference between the plates of a capacitor is 175 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant that the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.

The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about \(0.070 \mathrm{V}\) exists across the membrane. The thickness of the cell membrane is \(8.0 \times 10^{-9} \mathrm{m} .\) What is the magnitude of the electric field in the membrane?

A parallel plate capacitor has a capacitance of \(7.0 \mu \mathrm{F}\) whenfilled with a dielectric. The area of each plate is \(1.5 \mathrm{m}^{2}\) and the separation between the plates is \(1.0 \times 10^{-5} \mathrm{m} .\) What is the dielectric constant of the dielectric?

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