/*! 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 34 24.34. A 0.350 -m-long cylindric... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 34

24.34. A 0.350 -m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 \(\mathrm{mm}\) and an outer hollow conducting tube with an inner radius of 2.00 \(\mathrm{mm}\) . The two conductors are separated by air and charged to a potential difference of 6.00 \(\mathrm{V}\) . Calculate (a) the charge per length for the capacitor; (b) the total charge on the capacitor; (c) the capacitance; (d) the energy stored in the capacitor when fully charged.

Short Answer

Expert verified
Charge per length: \(1.4472 \times 10^{-10} \, \text{C/m}\). Total charge: \(5.0652 \times 10^{-11} \, \text{C}\). Capacitance: \(8.442 \times 10^{-12} \, \text{F}\). Energy stored: \(1.52 \times 10^{-10} \, \text{J}\).

Step by step solution

01

Calculate Capacitance per Unit Length

For a cylindrical capacitor, the capacitance per unit length \( C' \) is given by the formula: \[ C' = \frac{2\pi\varepsilon_0}{\ln\left(\frac{b}{a}\right)} \] where \( a = 1.20\, \text{mm} = 1.20 \times 10^{-3}\, \text{m} \) is the radius of the inner conductor, \( b = 2.00\, \text{mm} = 2.00 \times 10^{-3}\, \text{m} \) is the radius of the outer conductor, and \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) is the permittivity of free space. Substituting the values, we have: \[ C' = \frac{2\pi \times 8.85 \times 10^{-12}}{\ln\left(\frac{2.00 \times 10^{-3}}{1.20 \times 10^{-3}}\right)} \approx 2.412 \times 10^{-11} \, \text{F/m} \].
02

Calculate Charge per Unit Length

The charge per unit length \( \lambda \) is related to the capacitance per unit length by the formula: \[ \lambda = C' \cdot V \] where \( V = 6.00 \, \text{V} \) is the potential difference. Thus, \[ \lambda = (2.412 \times 10^{-11} \, \text{F/m}) \times 6.00 \, \text{V} \approx 1.4472 \times 10^{-10} \, \text{C/m} \].
03

Calculate Total Charge on the Capacitor

The total charge \( Q \) on the capacitor is calculated using: \[ Q = \lambda \times L \] where \( L = 0.350 \, \text{m} \) is the length of the capacitor. Thus, \[ Q = 1.4472 \times 10^{-10} \, \text{C/m} \times 0.350 \, \text{m} \approx 5.0652 \times 10^{-11} \, \text{C} \].
04

Calculate Total Capacitance

The total capacitance \( C \) of the capacitor is: \[ C = C' \times L \] Substituting the given values, we get \[ C = (2.412 \times 10^{-11} \, \text{F/m}) \times 0.350 \, \text{m} \approx 8.442 \times 10^{-12} \, \text{F} \].
05

Calculate Energy Stored in the Capacitor

The energy \( U \) stored in the capacitor is calculated using: \[ U = \frac{1}{2} C V^2 \] With \( C = 8.442 \times 10^{-12} \, \text{F} \) and \( V = 6.00 \, \text{V} \), the energy is: \[ U = \frac{1}{2} \times 8.442 \times 10^{-12} \times (6.00)^2 \approx 1.52 \times 10^{-10} \, \text{J} \].

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 Calculation
In order to calculate the capacitance of a cylindrical capacitor, it's crucial to understand that the capacitance is a measure of the ability of a capacitor to store charge per unit voltage applied.
The capacitance per unit length \(C'\) of a cylindrical capacitor is given by:
  • Using the formula: \[C' = \frac{2\pi\varepsilon_0}{\ln\left(\frac{b}{a}\right)}\]
Here, \( a \) is the radius of the inner cylinder, \( b \) is the radius of the outer cylinder, and \( \varepsilon_0 \) is the permittivity of free space, a constant value that indicates how much electric field is "allowed" in a particular space.
By substituting the given values into the equation, we can calculate the capacitance per unit length of the cylindrical capacitor. Understanding how these values come together helps in recognizing how different configurations of a capacitor can influence its capacity to store charge.
Electrostatics
Electrostatics is the branch of physics that deals with the study of forces, fields, and potentials arising from charged particles at rest. In the context of a cylindrical capacitor:
  • The potential difference \( V \) between two conductors creates an electric field \( E \).
  • The charge distributed on the surfaces follows the property of electrostatics that the field inside a conductor is zero in electrostatic equilibrium.
The electric field within the material of the cylindrical capacitor relates to how the charges are configured across the surfaces of the inner and outer conductors. The idea of electrostatics helps us to derive meaningful relationships between charge, electric field, and potential energy in such systems.
By setting the potential difference \( V \), we define the capacity of the capacitor to store more or less electric potential energy.
Energy Stored in Capacitors
Capacitors are known for their ability to store energy in the form of an electric field between their plates. The energy \( U \) stored in a cylindrical capacitor is determined by the capacitance and the square of the voltage\( V \) applied across it, shown by the formula:
  • \[ U = \frac{1}{2} C V^2 \]
This formula indicates that the energy stored is directly proportional to both the capacitance and the square of the voltage.
In practical terms, understanding this relationship helps in evaluating the efficiency and feasibility of the capacitor in various applications, such as filters in electronic circuits.
Calculating the exact energy stored also quantifies how efficiently a capacitor can release energy back into a circuit without effective power loss.

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

24.53. Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{6 / 5}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) . (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?

24.27. A \(450-\mu\) F capacitor is charged to 295 V. Then a wire is con- nected between the plates. How many joules of thermal energy are produced as the capacitor discharges if all of the energy that was stored goes into heating the wire?

24.48 . A parallel-plate capacitor has the volume between its plates filled with plastic with dielectric constant \(K\) . The magnitude of the charge on each plate is \(Q\) . Each plate has area \(A\) , and the distance between the plates is \(d\) . (a) Use Gauss's law as stated in Eq. \((24.23)\) to calculate the magnitude of the electric field in the

24.58. Several \(0.25-\mu \mathrm{F}\) capacitors are available. The voltage across each is not to exceed 600 \(\mathrm{V}\) . You need to make a capacitor with capacitance 0.25\(\mu \mathrm{F}\) to be connected across a potential difference of 960 \(\mathrm{V}\) . (a) Show in a diagram how an cquivalent capacitor with the desired properties can be obtained. (b) No dielectric is a perfect insulator that would not permit the flow of any charge through its volume. Suppose that the dielectric in one of the capacitors in your diagram is a moderately good conductor. What will happen in this case when your combination of capacitors is connected across the \(960-\mathbf{V}\) potential difference?

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?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Translational Dynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Physics of Motion

Read Explanation

Dynamics

Read Explanation

Work Energy and Power

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.