/*! 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 45 24.45. When a \(360-n F\) air ca... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 45

24.45. When a \(360-n F\) air capacitor \(\left(1 n F=10^{-9} \mathrm{F}\right)\) is connected to a power supply, the energy stored in the capacitor is \(1.85 \times 10^{-5}\) J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by \(2.32 \times 10^{-5} \mathrm{J}\) (a) What is the potential difference between the capacitor plates?(b) What is the dielectric constant of the slab?

Short Answer

Expert verified
(a) 10.14 V; (b) 2.28

Step by step solution

01

Find the initial potential difference

The energy stored in a capacitor can be expressed using the formula \( U = \frac{1}{2} C V^2 \), where \( U \) is the energy, \( C \) is the capacitance, and \( V \) is the potential difference. Given that \( U = 1.85 \times 10^{-5} \) J and \( C = 360 \times 10^{-9} \) F, we rearrange the formula to solve for \( V \):\[ V = \sqrt{\frac{2U}{C}} = \sqrt{\frac{2 \times 1.85 \times 10^{-5}}{360 \times 10^{-9}}} = \sqrt{102.777} \approx 10.14 \, \text{V}. \]Thus, the initial potential difference, before the dielectric is inserted, is approximately 10.14 V.
02

Calculate the total energy with the dielectric

When the dielectric is inserted, the total energy stored becomes \( 1.85 \times 10^{-5} \text{ J} + 2.32 \times 10^{-5} \text{ J} = 4.17 \times 10^{-5} \text{ J} \).
03

Find the capacitance with the dielectric

The formula for the energy stored in a capacitor remains \( U' = \frac{1}{2}C'V^2 \), where \( U' = 4.17 \times 10^{-5} \text{ J} \) is the new energy. From \( C' = \kappa C \), we substitute and get \( 4.17 \times 10^{-5} = \frac{1}{2}(\kappa C)V^2 \). Use \( V = 10.14 \text{ V} \) to solve for \( C' \):\[ \kappa C = \frac{2U'}{V^2} = \frac{2 \times 4.17 \times 10^{-5}}{(10.14)^2} \approx 820 \times 10^{-9} \text{ F}. \]The capacitance with the dielectric is approximately \( 820 \text{ nF} \).
04

Calculate the dielectric constant

Using the relationship \( C' = \kappa C \), where \( C' = 820 \text{ nF} \) and \( C = 360 \text{ nF} \), solve for \( \kappa \):\[ \kappa = \frac{C'}{C} = \frac{820}{360} \approx 2.28. \]Thus, the dielectric constant of the slab is approximately 2.28.

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.

Capacitor Energy
Capacitor energy refers to the energy stored within a capacitor, which is a device capable of storing electrical energy in an electric field. The energy stored in a capacitor can be calculated using the formula \( U = \frac{1}{2} C V^2 \), where:
  • \( U \) is the energy stored in joules (J),
  • \( C \) is the capacitance in farads (F),
  • \( V \) is the potential difference in volts (V).
The energy is a result of the work done to move charge onto the capacitor plates. When a dielectric material is introduced, it changes the capacitor's ability to store more energy. This happens because the dielectric reduces the electric field within the capacitor, allowing more charge to be stored for the same potential difference. This increase in charge (and thus energy) is directly related to the properties of the dielectric material used.
Dielectric Constant
The dielectric constant, also known as the relative permittivity, is a measure of the extent to which a dielectric material can concentrate electrostatic lines of flux. This constant is denoted by \( \kappa \). The presence of a dielectric affects the capacitance of a capacitor by a factor of the dielectric constant: \( C' = \kappa C \).
  • When \( \kappa > 1 \), it indicates the dielectric material increases the capacitance.
  • A higher \( \kappa \) value shows the material's greater ability to store electric charge.
  • The dielectric constant is dimensionless as it is a ratio.
Inserting a dielectric material leads to an increased total energy stored in the capacitor, as observed in the original problem, where a slab increased the stored energy by extending the capacitor's ability to hold more charge. Thus, the dielectric constant is crucial in enhancing capacitor performance.
Potential Difference
Potential difference, often referred to as voltage, is the work needed to move a unit charge between two points. In the context of capacitors, it's the difference in electric potential between the two plates of the capacitor. This potential difference is essential as it determines the amount of energy stored in the capacitor.
  • For a given capacitance, a higher potential difference results in more stored energy.
  • Measured in volts (V), it represents the electric "pressure" in the circuit.
  • A constant potential difference is often maintained across a capacitor using a power supply.
In capacitors with a dielectric, while the potential difference across the capacitor remains unchanged due to the power supply, the capacitance increases, allowing for greater energy storage. In the problem, the initial potential difference of 10.14 V was maintained even after inserting the dielectric, thereby increasing the energy stored by altering the capacitance instead.

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

24.41. The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of \(1.60 \times\) \(10^{7} \mathrm{V} / \mathrm{m}\) . The capacitor is to have a capacitance of \(1.25 \times 10^{-9} \mathrm{F}\) and must be able to withstand a maximum potential difference of 5500 \(\mathrm{V}\) . What is the minimum area the plates of the capacitor may have?

24.77. Three square metal plates \(A, B,\) and \(C,\) each 12.0 \(\mathrm{cm}\) on a side and 1.50 \(\mathrm{mm}\) thick, are arranged as in Fig. 24.43 . The plates are separated by sheets of paper 0.45 \(\mathrm{mm}\) thick and with dielectric constant \(4.2 .\) The outer plates are connected together and connected to point \(b\) . The inner plate is connected to point \(a\) (a) Copy the diagram and show by plus and minus signs the charge distribution on the plates when point \(a\) is maintained at a positive potential relative to point \(b\) . (b) What is the capacitance between points \(a\) and \(b ?\)

24.66. An air capacitor is made by using two flat plates, each with area \(A\) , separated by a distance \(d\) . Then a metal slab having thickness a (less than \(d\) ) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (Fig. 24.36\()\) . (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance \(C_{0}\) when the metal slab is not present.(c) Discuss what happens to the capacitance in the limits \(a \rightarrow 0\) and \(a \rightarrow d\) .

24.26. An air capacitor is made from two flat parallel plates 1.50 \(\mathrm{mm}\) apart. The magnitude of charge on each plate is 0.0180\(\mu \mathrm{C}\) when the potential difference is 200 \(\mathrm{V}\) . (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m}\) . (d) When the charge is 0.0180\(\mu \mathrm{C}\) , what total energy is stored?

24.20. Two parallel plate vacuum capacitors have plate spacings \(d_{1}\) and \(d_{2}\) and equal plate areas \(A\) . Show that when the capacitors are connected in series, the eqnivalent capacitance is the same as for a single capacitor with plate area \(A\) and spacing \(d_{1}+d_{2}\) .
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Measurements

Read Explanation

Oscillations

Read Explanation

Circular Motion and Gravitation

Read Explanation

Linear Momentum

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.