/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 In the text, it was shown that t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the text, it was shown that the energy stored in a capacitor \(C\) charged to a potential \(V\) is \(U=\frac{1}{2} Q V .\) Show that this energy can also be expressed as (a) \(U=Q^{2} / 2 C\) and (b) \(U=\frac{1}{2} C V^{2}\).

Short Answer

Expert verified
The energy can be expressed as \(U=Q^{2}/2C\) and \(U=\frac{1}{2}CV^{2}\).

Step by step solution

01

Review the Energy Formula

We start with the energy stored in a capacitor which is given by the formula \( U = \frac{1}{2} Q V \). This is the starting point for our transformations.
02

Express Voltage in Terms of Charge and Capacitance

Recall the relation between voltage \( V \), charge \( Q \), and capacitance \( C \): \( V = \frac{Q}{C} \).
03

Substitute Voltage in Energy Formula for Part (a)

Substitute \( V = \frac{Q}{C} \) into the original energy formula: \[ U = \frac{1}{2} Q \left(\frac{Q}{C}\right) = \frac{1}{2} \frac{Q^2}{C}. \] This confirms part (a) of the exercise, showing that \( U = \frac{Q^2}{2C} \).
04

Express Charge in Terms of Voltage and Capacitance

Using the same relation, express \( Q \) in terms of \( V \) and \( C \): \( Q = CV \).
05

Substitute Charge in Energy Formula for Part (b)

Substitute \( Q = CV \) into the original energy formula: \[ U = \frac{1}{2} (CV)V = \frac{1}{2} CV^2. \] This confirms part (b) of the exercise, showing that \( U = \frac{1}{2} CV^2 \).

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 91Ó°ÊÓ!

Key Concepts

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

Capacitance
Capacitance is a fundamental concept in electrostatics that measures a capacitor's ability to store electric charge. Essentially, it represents how much charge a capacitor can hold when a specific electric potential (voltage) is applied across its terminals. Capacitance is denoted by the symbol \[ C \] and is measured in farads (F). It can be calculated using the formula:\[C = \frac{Q}{V}\]where
  • \(Q\) is the charge stored in the capacitor, and
  • \(V\) is the electric potential difference.
A higher capacitance indicates that a capacitor can store more charge at the same voltage, which is crucial in various electronic and electrical applications, such as smoothing voltage fluctuations in power supplies.
Electric Potential
Electric potential, often referred to as voltage, is a measure of the potential energy per unit charge at a point in space due to an electric field. It indicates how much work is required to move a unit charge from a reference point to a specified point without any acceleration.The potential difference (voltage) between two points is represented by \[ V \] and is measured in volts (V). In the context of capacitors, electric potential helps to determine how much energy is stored within the capacitor. This relationship can be mathematically expressed using the formula:\[ Q = CV \]where
  • \(Q\) is the charge (in coulombs),
  • \(C\) is the capacitance, and
  • \(V\) is the voltage.
Understanding electric potential is key to analyzing circuits and understanding how charges and energy move within them.
Electrostatics
Electrostatics is the study of electric charges at rest. It involves understanding how electric forces act between charged objects, as well as how charges can be distributed and influenced by electric fields.One essential principle in electrostatics is Coulomb's law, which defines the force between two charges:\[ F = k_e \frac{|q_1 q_2|}{r^2} \]where
  • \(F\) is the force between the charges,
  • \(q_1\) and \(q_2\) are the magnitudes of the charges,
  • \(r\) is the distance between them, and
  • \(k_e\) is Coulomb's constant.
Electrostatics plays a significant role in understanding capacitance because it describes how charges are stored on the plates of a capacitor due to the applied electric field. The separation of these charges is what allows the capacitor to store electrical energy.
Energy Transformation
Energy transformation in capacitors is the process by which electrical energy is stored in an electric field and later used to do work. The stored energy can be calculated using different expressions depending on known quantities, such as charge and voltage.The primary formula for energy stored in a capacitor is:\[ U = \frac{1}{2} QV \]However, by substituting the basic relationships between charge \(Q\), voltage \(V\), and capacitance \(C\), this expression can be rewritten:
  • In terms of charge and capacitance: \(U = \frac{Q^2}{2C}\)
  • In terms of voltage and capacitance: \(U = \frac{1}{2} CV^2\)
Energy transformation is vital for devices like flash cameras and defibrillators, which release stored energy when needed. Understanding these energy interactions helps engineers design efficient systems that store and deliver electrical energy effectively.

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 point charge \(Q=+4.60 \mu \mathrm{C}\) is held fixed at the origin. A second point charge \(q=+1.20 \mu \mathrm{C}\) with mass of \(2.80 \times 10^{-4} \mathrm{~kg}\) is placed on the \(x\) axis, \(0.250 \mathrm{~m}\) from the origin. (a) What is the electric potential energy \(U\) of the pair of charges? (Take \(U\) to be zero when the charges have infinite separation.) (b) The second point charge is released from rest. What is its speed when its distance from the origin is (i) \(0.500 \mathrm{~m} ;\) (ii) \(5.00 \mathrm{~m} ;\) (iii) \(50.0 \mathrm{~m} ?\)

X-ray tube. An X-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the time they reach the other end. The acceleration is accomplished using an electric field. The high-speed electrons hit a metal target at the other end, and the violence of the collision converts their kinetic energy into high-energy light rays, commonly known as X-rays. (a) Through what potential difference should electrons be accelerated so that their speed is \(1.0 \%\) of the speed of light when they hit the target? (b) What potential difference would be needed to give protons the same kinetic energy as the electrons? (c) What speed would this potential difference give to protons? Express your answer in \(\mathrm{m} / \mathrm{s}\) and as a percent of the speed of light.

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{~m}\) ) and point \(b\) (at \(x=0.90 \mathrm{~m}\) ) is \(240 \mathrm{~V}\). (a) Which point, \(a\) or \(b\), is at the higher potential? (b) Calculate the value of \(E\). (c) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to \(a\). Calculate the work done on the point charge by the electric field.

DATA A positive point charge \(Q\) is placed at a position \(x_{0}\) on the \(x\) axis. The potential \(V\) is then measured at various points along the \(x\) axis. The resulting data are given in the table: $$ \begin{array}{ll} \hline x \text { position (m) } & \text { Potential }(V) \\ \hline 0 & 90 \\ 0.5 & 45 \\ 1.5 & 23 \\ 2.0 & 18 \\ 2.5 & 15 \\ \hline \end{array} $$ Make a plot of \(1 / V\) as a function of the position. From this plot, determine the position and magnitude of the charge.

Two very large metal parallel plates that are \(25 \mathrm{~cm}\) apart, oriented perpendicular to a sheet of paper, are connected across the terminals of a \(50.0 \mathrm{~V}\) battery. (a) Draw to scale the lines where the equipotential surfaces due to these plates intersect the paper. limit your drawing to the region between the plates, avoiding their edges, and draw the lines for surfaces that are \(10.0 \mathrm{~V}\) apart, starting at the lowpotential plate. (b) These surfaces are separated equally in potential. Are they also separated equally in distance? (c) In words, describe the shape and orientation of the surfaces you just found.

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.