/*! 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 53 A parallel-plate vacuum capacito... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 53

A parallel-plate vacuum capacitor has \(8.38 \mathrm{~J}\) of energy stored in it. The separation between the plates is \(2.30 \mathrm{~mm} .\) If the separation is decreased to \(1.15 \mathrm{~mm},\) what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

Short Answer

Expert verified
(a) Energy is 4.19 J (constant charge), (b) Energy is 16.76 J (constant voltage).

Step by step solution

01

Understanding Initial Setup

A parallel-plate capacitor is given with an initial energy of \(8.38 \, \text{J}\) and a plate separation of \(2.30 \, \text{mm}\). We need to find the stored energy when the plate separation changes under two different conditions.
02

Energy with Constant Charge (Disconnected)

When a capacitor is disconnected, the charge \(Q\) remains constant. The energy stored in a capacitor is given by the formula: \[ U = \frac{1}{2} \frac{Q^2}{C} \]Since \(U \propto C^{-1}\), and capacitance \(C\) is inversely proportional to the distance \(d\), if \(d\) is halved, \(C\) doubles. Thus, the new energy \(U'\) will be half of the original energy. Therefore, \[ U' = \frac{1}{2} \times 8.38 \, \text{J} = 4.19 \, \text{J} \]
03

Energy with Constant Voltage (Connected)

When a capacitor remains connected to a voltage source, the potential difference \(V\) stays constant. In this case, the energy stored is given by:\[ U = \frac{1}{2} C V^2 \]Here, since \(C\) doubles (as \(d\) is halved), the energy \(U'\) will also double. Therefore, \[ U' = 2 \times 8.38 \, \text{J} = 16.76 \, \text{J} \]

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!

Key Concepts

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

Capacitance
Capacitance is a fundamental property of capacitors that describes their ability to store electric charge. It is denoted by the symbol \(C\). For a parallel-plate capacitor, capacitance is influenced by the area of the plates \(A\), the distance between them \(d\), and the permittivity of the medium between the plates \(\varepsilon\). The formula to calculate capacitance is given by:\[ C = \frac{\varepsilon A}{d} \]The unit of capacitance is the farad (F).
  • An increase in plate area \(A\) or a decrease in the distance \(d\) increases capacitance, meaning the capacitor can store more charge.
  • When the distance between the plates is reduced, the capacitance increases.
Understanding capacitance is crucial for analyzing how capacitors behave when conditions change, such as when they are connected or disconnected from a potential source.
Stored Energy
The energy stored in a capacitor is the energy required to maintain a charge separation on the plates. This energy is proportional to the capacitance \(C\) and the square of the voltage \(V\) across the plates, described by the formula:\[ U = \frac{1}{2} C V^2 \]Alternatively, when the charge \(Q\) and capacitance are known, the energy is given by:\[ U = \frac{1}{2} \frac{Q^2}{C} \]The stored energy can change based on whether the capacitor is connected to a power source or not:
  • If connected, the voltage remains constant, and a decrease in plate distance doubles the energy, as capacitance doubles.
  • If disconnected, the charge remains constant, and energy halves because capacitance inversely affects stored energy.
Voltage
Voltage, also known as potential difference, is a measure of the electric potential energy difference between two points, such as the plates of a capacitor. In a capacitor, voltage influences the amount of charge it can hold for a given capacitance. The relationship is given by:\[ V = \frac{Q}{C} \]Maintaining a constant voltage means sustaining a consistent potential difference across the capacitor's plates. If a capacitor stays connected to a battery:
  • The voltage across its plates does not change, regardless of changes in plate distance.
  • The energy stored will adjust according to the changes in capacitance.
In scenarios where voltage stays stable, it crucially directs how stored energy responds to changes in physical configuration.
Capacitor Disconnected
When a capacitor is disconnected from its voltage source, it operates under constant charge conditions. This means that the total charge \(Q\) stored on the plates does not change even if the plate separation changes.
  • The capacitance may change if the plate separation is altered, but since the charge is fixed, the effect on stored energy is distinct.
  • The stored energy \(U\) depends on capacitance according to the formula \( U = \frac{1}{2} \frac{Q^2}{C} \), so if capacitance increases, stored energy decreases.
In the given scenario, when the plate separation is halved, capacitance doubles, and the stored energy is halved. This results in 4.19 J of stored energy when the capacitor is disconnected.
Capacitor Connected
For a capacitor that remains connected to a potential source, the situation changes significantly. The voltage \(V\) across the capacitor stays constant because the circuit with the voltage source remains intact.
  • When the separation between plates is reduced, the capacitance increases, since \(C\) is inversely related to the distance \(d\).
  • According to the formula for energy \( U = \frac{1}{2} C V^2 \), a doubling in capacitance results in doubling the stored energy.
This is why, in the given exercise, reducing the plate separation while keeping the capacitor connected results in the energy increasing to 16.76 J. The constant voltage ensures that energy changes in direct proportion to changes in capacitance.

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

\(\mathrm{A} 12.5 \mu \mathrm{F}\) capacitor is connected to a power supply that keeps a constant potential difference of \(24.0 \mathrm{~V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

For a particular experiment, helium ions are to be given a kinetic energy of \(3.0 \mathrm{MeV}\). What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? A. \(-3.0 \mathrm{MV}\) B. \(+3.0 \mathrm{MV}\) C. \(+1.5 \mathrm{MV}\) D. \(+1.0 \mathrm{MV}\)

Two large metal parallel plates carry opposite charges of equal magnitude. They are separated by \(45.0 \mathrm{~mm},\) and the potential difference between them is \(360 \mathrm{~V}\). (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+2.40 \mathrm{nC} ?\)

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) \(50.0 \mathrm{~m} ?\)

A helium ion \(\left(\mathrm{He}^{++}\right)\) that comes within about \(10 \mathrm{fm}\) of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of \(3.0 \mathrm{MeV}\) heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than \(10 \mathrm{fm}\) from the center of the atomic nucleus? (1 \(\mathrm{fm}=1 \times 10^{-15} \mathrm{~m},\) and \(e\) is the magnitude of the charge of an electron or a proton. A. \(2 e\) B. \(11 e\) C. \(20 e\) D. \(22 e\)
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Rotational Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation

Astrophysics

Read Explanation

Work Energy and Power

Read Explanation

Engineering 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.