/*! 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 26 An air capacitor is made from tw... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
(a) Capacitance is \(9.0 \times 10^{-11} \text{ F}\). (b) Area is \(0.0152 \text{ m}^2\). (c) Maximum voltage is 4500 V. (d) Stored energy is \(1.8 \times 10^{-6} \text{ J}\).

Step by step solution

01

Calculate Capacitance

Capacitance is given by the formula \( C = \frac{Q}{V} \), where \( Q \) is the charge and \( V \) is the potential difference. Substituting the given values:\[ C = \frac{0.0180 \times 10^{-6} \text{ C}}{200 \text{ V}} = 9.0 \times 10^{-11} \text{ F} \]
02

Calculate Plate Area

The capacitance for a parallel plate capacitor is also given by \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation. Rearranging for \( A \):\[ A = \frac{C \times d}{\varepsilon_0} \]\[ A = \frac{9.0 \times 10^{-11} \text{ F} \times 1.50 \times 10^{-3} \text{ m}}{8.85 \times 10^{-12} \text{ F/m}} = 0.0152 \text{ m}^2 \]
03

Calculate Maximum Voltage

The maximum voltage \( V_{\text{max}} \) before dielectric breakdown is given by the maximum electric field \( E \) times the separation \( d \):\[ V_{\text{max}} = E \times d = 3.0 \times 10^{6} \text{ V/m} \times 1.50 \times 10^{-3} \text{ m} = 4500 \text{ V} \]
04

Calculate Stored Energy

The energy \( U \) stored in a capacitor is given by \( U = \frac{1}{2}CV^2 \). Substituting the given values:\[ U = \frac{1}{2} \times 9.0 \times 10^{-11} \text{ F} \times (200 \text{ V})^2 = 1.8 \times 10^{-6} \text{ J} \]

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 91Ó°ÊÓ!

Key Concepts

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

Parallel Plate Capacitor
A parallel plate capacitor consists of two conductive plates facing each other, separated by a small distance. The capacitor stores electrical energy through the separation of charges on these plates. The capacitance, measured in farads (F), is the ability of a capacitor to store an electric charge for a given potential difference. In this arrangement, when a voltage is applied, one plate accumulates positive charge while the other accumulates negative charge. The formula for capacitance in this context is given by \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between them.
  • Key factors affecting capacitance: Larger plate area increases capacitance, as there is more surface for charge storage.
  • Smaller distance between plates increases capacitance, enabling stronger electrical interaction between the charges on the plates.
Understanding these principles is fundamental to effectively work with parallel plate capacitors in various electronic applications.
Dielectric Breakdown
Dielectric breakdown is a phenomenon that occurs when a material (insulator) becomes conductive. In the case of capacitors, if the electric field between plates becomes too strong, it can cause the insulating material (often air, as in this problem) to conduct electricity. This results in a short circuit and damage to the capacitor.
  • Maximum electric field: For air, the dielectric breakdown occurs at electric field strengths around \(3.0 \times 10^6 \text{ V/m}\).
  • Avoidance: Ensuring the voltage across the capacitor does not exceed this threshold helps in preventing breakdown.
Thus, calculating the maximum voltage a capacitor can handle without undergoing breakdown is crucial.
Electric Field Strength
Electric field strength within a parallel plate capacitor is a measure of the force a charge experiences within the field. It is defined as the voltage between the plates divided by the distance separating them: \( E = \frac{V}{d} \).
  • Uniform field: Between parallel plates, the electric field is typically uniform, meaning its strength and direction remain constant.
  • Dependence on voltage and separation: Higher voltages or smaller separations increase the field strength, increasing the potential for dielectric breakdown.
Understanding the electric field's behavior helps in designing capacitors that can effectively cope with high voltages without failure.
Stored Energy in Capacitors
The stored energy in a capacitor is given by the formula \( U = \frac{1}{2}CV^2 \), where \( C \) is capacitance and \( V \) is the voltage. This energy is the work needed to establish both the voltage and charge in the capacitor. It acts as a reservoir which can be released to do useful work in electronic circuits.
  • Importance in circuits: Capacitors are used in various applications to maintain power supply, filter signals, and stabilize voltage and power flow.
  • Energy storage capability: Depends on both the capacitance and the square of the voltage applied, thus emphasizing the critical role of both factors.
With this understanding, the potential utility and limitations of capacitors can be realistically estimated in practical scenarios.

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

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, air-filled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 42.0 \(\mathrm{mm}^{2}\) , and the separation between the plates is 0.700 \(\mathrm{mm}\) before the key is depressed. (a) Calculate the capacitance before the key is depressed. (b) If the circulatry can detect a change in capacitance of 0.250 \(\mathrm{pF}\) , how far must the key be depressed before the circuitry detects its depression?

Capacitance of a Thundercloud. The charge center of a thundercloud, drifting 3.0 \(\mathrm{km}\) above the earth's surface, contains 20 \(\mathrm{C}\) of negative charge. Assuming the charge center has a radius of \(1.0 \mathrm{km},\) and modeling the charge center and the earth's surface as parallel plates, calculate: (a) the capacitance of the system; (b) the potential difference between charge center and ground; (c) the average strength of the electric field between cloud and ground; (d) the electrical energy stored in the system.

A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148\(\mu \mathrm{C}\) on each plate. The plates are 0.328 \(\mathrm{mm}\) apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric- field magnitude between the plates? (d) What is the surface charge density on each plate?

A \(10.0-\mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a 12.0 -V battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the \(12.0-\mathrm{V}\) battery after the radius of each plate was doubled without changing their separation?

The plates of a parallel-plate capacitor are 3.28 \(\mathrm{mm}\) apart, and each has an area of 12.2 \(\mathrm{cm}^{2} .\) Each plate carries a charge of magnitude \(4.35 \times 10^{-8} \mathrm{C}\) . The plates are in vacuum. (a) What is the capacitance? (b) What is the potential difference between the plates? (c) What is the magnitude of the electric field between the plates?

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.