/*! 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 31 A cylindrical air capacitor of l... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A cylindrical air capacitor of length 15.0 m stores 3.20 \(\times\) 10\(^{-9}\) J of energy when the potential difference between the two conductors is 4.00 V. (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

Short Answer

Expert verified
(a) Charge on each conductor is 1.60 × 10^{-9} C. (b) Ratio of radii is approximately 1.0868.

Step by step solution

01

Use Energy Formula for Capacitors

The energy stored in a capacitor is given by the formula \[U = \frac{1}{2} C V^2\]where \( U \) is the energy, \( C \) is the capacitance, and \( V \) is the potential difference. We can rearrange this formula to solve for capacitance \( C \):\[C = \frac{2U}{V^2}\]
02

Calculate Capacitance

Substitute the given values into the formula:\[C = \frac{2 \times 3.20 \times 10^{-9}}{4.00^2}\]Calculate C:\[C = \frac{6.40 \times 10^{-9}}{16} = 4.00 \times 10^{-10} \text{ F}\]
03

Use Capacitance Formula to Find Charge

The charge \( Q \) on the conductors is given by:\[Q = CV\]Substituting the known values:\[Q = (4.00 \times 10^{-10} \text{ F})(4.00 \text{ V})\]Calculate \( Q \):\[Q = 1.60 \times 10^{-9} \text{ C}\]
04

Use Capacitance Formula for Cylinders

The capacitance \( C \) of a cylindrical capacitor is given by:\[C = \frac{2\pi \varepsilon_0 L}{\ln\left(\frac{b}{a}\right)}\]where \( L \) is the length, \( a \) is the radius of the inner cylinder, \( b \) is the radius of the outer cylinder, and \( \varepsilon_0 \) is the vacuum permittivity \((8.85 \times 10^{-12} \text{ F/m})\).
05

Solve for the Ratio of Radii

From the given formula:\[\frac{b}{a} = e^{\left(\frac{2\pi \varepsilon_0 L}{C}\right)}\]Substitute the values for calculation:\[\frac{b}{a} = e^{\left(\frac{2\pi (8.85 \times 10^{-12}) (15.0)}{4.00 \times 10^{-10}}\right)}\]Calculate inside the exponent and then solve the exponent:\[\frac{b}{a} = e^{0.083387}\]\[\frac{b}{a} \approx 1.0868\]
06

Conclusion

With the charge calculated as \(1.60 \times 10^{-9} \text{ C}\) and the ratio of radii approximately \( 1.0868 \), the exercise is complete.

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 property of a capacitor. It reflects a capacitor's ability to store an electric charge per unit of electrical potential difference across its conductors. Mathematically, it is represented as \( C = \frac{Q}{V} \),where \( C \) is the capacitance in farads (F), \( Q \) is the charge in coulombs (C), and \( V \) is the potential difference in volts (V).
For different shapes and materials, capacitors have unique formulas for calculating capacitance, but the principle remains the same: it measures how effectively a capacitor can store electrical energy given a certain voltage. In cylindrical capacitors, the capacitance also depends on their physical dimensions, such as length, as well as the radii of the conducting surfaces.
Understanding capacitance is essential because it helps predict how much energy a capacitor can store, making it crucial for designing electrical circuits and systems.
Charge Calculation
Calculating the charge on a capacitor involves understanding the relationship between charge, capacitance, and voltage. The charge \( Q \) stored in a capacitor with a capacitance \( C \) and a voltage \( V \) applied across it is given by:\[ Q = C \times V \]This formula implies that the charge stored is directly proportional to both the capacitance and the voltage.
In the context of our problem, with a given capacitance of \( 4.00 \times 10^{-10} \) F (farads) and a potential difference of 4.00 V, calculating the charge requires just simple multiplication:\[ Q = (4.00 \times 10^{-10} \, \text{F})(4.00 \, \text{V}) = 1.60 \times 10^{-9} \, \text{C} \]This equation highlights the key mechanism of energy storage in capacitors by means of charge accumulation between its plates. It’s a straightforward but powerful concept that underscores much of capacitor functionality in electronics.
Potential Difference
The potential difference, or voltage, across a capacitor is a critical factor that determines the amount of charge and energy it can store. It represents the work needed to move a unit charge between two points in an electric field.
In simple terms, potential difference is what pushes electrons through a circuit and into the capacitor’s plates. It's calculated as:\[ V = \frac{U}{Q} \],connecting energy \( U \) and charge \( Q \).
In our example, you are given a potential difference of 4.00 V. This value is essential for calculating both the charge on the capacitor and ultimately the energy stored. To maintain safety and efficiency, especially in practical applications, it’s important that capacitors are not subjected to voltages higher than their rated capacity to prevent the risk of damage or failure.
Energy Stored in Capacitor
The energy stored in a capacitor is determined by the formula:\[ U = \frac{1}{2} C V^2 \]This equation shows that the energy \( U \) is dependent on both the capacitance \( C \) and the square of the potential difference \( V \).
This relationship demonstrates how even a small increase in voltage can significantly increase the energy stored, making voltage levels critical in capacitor design.
In this exercise, the energy already stored is 3.20 \( \times \) 10\(^{-9}\) J. With this information, we can confirm the capacitor's effectiveness by ensuring that the energy calculation aligns with expected performance parameters. It’s crucial for applications in power supply systems, where accurate energy storage predictions contribute to stable and reliable power delivery.
Vacuum Permittivity
Vacuum permittivity, denoted as \( \varepsilon_0 \), is a constant value representing the ability of a vacuum to permit electric field lines. It is a critical factor in calculating the capacitance of various capacitor configurations.
For a cylindrical capacitor, the capacitance is given by:\[ C = \frac{2\pi \varepsilon_0 L}{\ln\left(\frac{b}{a}\right)} \],where \( L \) is the length, \( a \) is the radius of the inner cylinder, and \( b \) is the radius of the outer cylinder.
\( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \text{ F/m} \) (farads per meter). This constant is fundamental in the laws that govern electromagnetism, describing how electric fields interact in free space.
By influencing capacitance, vacuum permittivity affects how capacitors are designed and how they function in electrical circuits, playing a decisive role in innovations within the electronics industry.

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 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00 \(\times\) 10\(^2\) V. The electric field between the plates is to be no greater than 1.00 \(\times\) 10\(^4\) N/C. As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

A 12.5-\(\mu\)F capacitor is connected to a power supply that keeps a constant potential difference of 24.0 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?

A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

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 \({1 \over 675}\) s with an average light power output of 2.70 \(\times\) 10\(^5\) 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 V when the stored energy equals the value calculated in part (a). What is the capacitance?

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E = 3.20\times10^5\space V/m\). When the space is filled with dielectric, the electric field is \(E = 2.50\times10^5\space V/m\). (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

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.