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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?

Short Answer

Expert verified
The potential difference is approximately 1103 V.

Step by step solution

01

Understand the Formula

The electric potential energy (U) stored in a capacitor is related to its capacitance (C) and the potential difference (V) across it by the formula: \[ U = \frac{1}{2} C V^2 \]. We need to find V, the potential difference.
02

Rearrange the Formula

To find the potential difference (V), we need to rearrange the formula: \[ V^2 = \frac{2U}{C} \] and then find \( V \) by taking the square root.
03

Convert Units

Ensure that the capacitance is in Farads. Here, \( 120 \mu \mathrm{F} = 120 \times 10^{-6} \mathrm{F} \).
04

Insert Values

Insert the given values into the formula: \[ V^2 = \frac{2 \times 73}{120 \times 10^{-6}} \].
05

Calculate V^2

Calculate \( V^2 \) by performing the division: \[ V^2 = \frac{146}{120 \times 10^{-6}} = \frac{146}{0.00012} = 1216666.67 \].
06

Find V

Calculate the potential difference by taking the square root of \( V^2 \): \[ V = \sqrt{1216666.67} \approx 1103 \mathrm{V} \].

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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 understanding how capacitors operate. A capacitor is a device that stores electrical energy, with its ability to hold charge measured by its capacitance. This property is expressed as the ratio of the electric charge (Q) stored on each plate of the capacitor to the potential difference (V) across the plates. It is symbolized by the letter \( C \) and its unit is the Farad (F). In simpler terms, capacitance tells us how much charge a capacitor can store for a given voltage.
A larger capacitance means a capacitor can store more charge for the same voltage. In the exercise, a capacitance of \( 120 \mu \mathrm{F} \) means the capacitor can store \(120 \times 10^{-6} \mathrm{Coulombs} \) of charge per volt across its plates. This parameter becomes crucial when calculating other factors like potential difference and stored energy.
Potential Difference
Potential difference, also known as voltage, is the difference in electric potential between two points in a circuit. It is what drives the flow of charge in a circuit, acting like a pressure that pushes electrons around. In terms of a capacitor, potential difference is the voltage across the two plates of the capacitor. It is influenced by the amount of charge stored on the plates and the capacitor's capacitance.
The potential difference is crucial because it indicates how much energy per charge is stored in the capacitor. A higher potential difference across a capacitor typically means that the capacitor is storing more energy. In our exercise, the voltage or potential difference was found to be approximately \(1103 \mathrm{V}\), calculated from the given energy and capacitance values using the potential energy formula.
Capacitor Formula
The capacitor formula is key in understanding the relationship between energy, capacitance, and potential difference in a capacitor. The formula given as \( U = \frac{1}{2} C V^2 \) links these three critical pieces.
This equation shows that the electric potential energy \( U \), stored in a capacitor, is directly proportional to the capacitance \( C \) and the square of the potential difference \( V \). To find any of these quantities, you can rearrange the formula accordingly.
For example, to find \( V \) from the exercise, you rearrange to \( V^2 = \frac{2U}{C} \), and then solve for \( V \) by taking the square root. It's important to ensure all units are compatible—here, converting microfarads to farads was necessary.
This formula underlines the physical insights and calculations in problems involving capacitors, enabling us to determine how changes in one quantity (capacitor size, stored energy, or voltage) affect the others.

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Most popular questions from this chapter

Equipotential surface \(A\) has a potential of \(5650 \mathrm{V},\) while equipotential surface \(B\) has a potential of 7850 V. A particle has a mass of \(5.00 \times\) \(10^{-2} \mathrm{kg}\) and a charge of \(+4.00 \times 10^{-5} \mathrm{C} .\) The particle has a speed of \(2.00 \mathrm{m} / \mathrm{s}\) on surface \(A\). A nonconservative outside force is applied to the particle, and it moves to surface \(B\), arriving there with a speed of \(3.00 \mathrm{m} / \mathrm{s}\). How much work is done by the outside force in moving the particle from \(A\) to \(B ?\)

A charge of \(+125 \mu \mathrm{C}\) is fixed at the center of a square that is \(0.64 \mathrm{m}\) on a side. How much work is done by the electric force as a charge of \(+7.0 \mu \mathrm{C}\) is moved from one corner of the square to any other empty corner? Explain.

One particle has a mass of \(3.00 \times 10^{-3} \mathrm{kg}\) and a charge of \(+8.00 \mu \mathrm{C}\). A second particle has a mass of \(6.00 \times 10^{-3} \mathrm{kg}\) and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is \(0.100 \mathrm{m},\) the speed of the \(3.00 \times 10^{-3} \mathrm{kg}\) particle is \(125 \mathrm{m} / \mathrm{s} .\) Find the initial separation between the particles.

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?

Two identical capacitors store different amounts of energy: capacitor A stores \(3.1 \times 10^{-3} \mathrm{J},\) and capacitor \(\mathrm{B}\) stores \(3.4 \times 10^{-4} \mathrm{J} .\) The voltage across the plates of capacitor B is 12 V. Find the voltage across the plates of capacitor A.

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