/*! 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 80 The capacitor of Problem 79 is i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 80

The capacitor of Problem 79 is initially charged to a \(150-\mathrm{V}\) potential difference. The plates are then physically separated by another \(0.750 \mathrm{mm}\) in such a way that none of the charge can leak off the plates. Find (a) the new capacitance and (b) the new energy stored in the capacitor. Explain the result using conservation of energy.

Short Answer

Expert verified
(b) What is the new energy stored in the capacitor after the additional separation?

Step by step solution

01

Recall the capacitance formula for a parallel plate capacitor

The capacitance of a parallel plate capacitor can be calculated using the following formula: $$C=\frac{\epsilon A}{d}$$ where \(C\) is the capacitance, \(\epsilon\) is the vacuum permittivity constant, \(A\) is the area of one of the capacitor plates, and \(d\) is the distance between the plates.
02

Calculate the initial capacitance

Initially, the capacitor is charged to a potential difference of 150 V, and we know its capacitance from Problem 79 (which we denote as \(C_{1}\)). Given the potential difference \(V_{1}\) and the charge \(Q\) (which remains constant when the plates are separated), we can calculate the initial energy stored in the capacitor using the formula: $$U_{1}=\frac{1}{2}C_{1}V_{1}^2$$
03

Calculate the new distance between the plates

The plates of the capacitor are separated by an additional 0.750 mm, so we need to calculate the new distance between them, denoted as \(d_{2}\). Let \(d_{1}\) be the initial distance between the plates. We have: $$d_{2}=d_{1}+0.750\times10^{-3}$$
04

Calculate the new capacitance

To find the new capacitance, \(C_{2}\), we use the formula for the capacitance of a parallel plate capacitor, with the new distance \(d_2\) calculated in the previous step: $$C_{2}=\frac{\epsilon A}{d_{2}}$$
05

Calculate the new potential difference

We know that the charge on the capacitor remains constant. Using the relationship between charge, capacitance, and potential difference, \(Q=C_1V_1=C_2V_2\), we can calculate the new potential difference, \(V_2\): $$V_{2}=\frac{C_{1}}{C_{2}}V_{1}$$
06

Calculate the new energy stored in the capacitor

We can now find the new energy stored in the capacitor, \(U_2\), using the new capacitance, \(C_2\), and the new potential difference, \(V_2\). The formula for the energy stored in a capacitor is: $$U_{2}=\frac{1}{2}C_{2}V_{2}^2$$
07

Explain the result using conservation of energy

According to the conservation of energy principle, the initial energy stored in the capacitor should be equal to the new energy stored plus the work done to separate the plates. Mathematically, this can be written as: $$U_{1}=U_{2}+W$$ Where \(W\) is the work done to separate the plates. This can be used to analyze the results obtained in the previous steps and to verify the conservation of energy in this case.

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 91影视!

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

Suppose you were to wrap the Moon in aluminum foil and place a charge \(Q\) on it. What is the capacitance of the Moon in this case? [Hint: It is not necessary to have two oppositely charged conductors to have a capacitor. Use the definition of potential for a spherical conductor and the definition of capacitance to get your answer. \(]\)
An electron is accelerated from rest through a potential difference $\Delta V\(. If the electron reaches a speed of \)7.26 \times 10^{6} \mathrm{m} / \mathrm{s},$ what is the potential difference? Be sure to include the correct sign. (Does the electron move through an increase or a decrease in potential?)
A spherical conductor of radius \(R\) carries a total charge Q. (a) Show that the magnitude of the electric field just outside the sphere is \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the charge per unit area on the conductor's surface. (b) Construct an argument to show why the electric field at a point \(P\) just outside any conductor in electrostatic equilibrium has magnitude \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the local surface charge density. [Hint: Consider a tiny area of an arbitrary conductor and compare it to an area of the same size on a spherical conductor with the same charge density. Think about the number of field lines starting or ending on the two areas.]
An electron (charge \(-e\) ) is projected horizontally into the space between two oppositely charged parallel plates. The electric field between the plates is \(500.0 \mathrm{N} / \mathrm{C}\) upward. If the vertical deflection of the electron as it leaves the plates has magnitude \(3.0 \mathrm{mm},\) how much has its kinetic energy increased due to the electric field? [Hint: First find the potential difference through which the electron moves.]
A point charge \(q=+3.0\) nC moves through a potential difference $\Delta V=V_{\mathrm{f}}-V_{\mathrm{i}}=+25 \mathrm{V} .$ What is the change in the electric potential energy?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Electromagnetism

Read Explanation

Oscillations

Read Explanation

Magnetism

Read Explanation

Electricity and Magnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
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.