Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 \(\mu\)C. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

Short Answer

Expert verified
(a) 4.24 kV, (b) 8.48 kV, (c) 8.28 mJ

Step by step solution

01

Understanding Capacitance and Potential Difference

The relationship between charge \( Q \), capacitance \( C \), and potential difference \( V \) is given by the equation: \[ V = \frac{Q}{C} \]We are given \( C = 920 \times 10^{-12} \text{ F} \) (convert from pF to F) and \( Q = 3.90 \times 10^{-6} \text{ C} \) (convert from \( \mu \text{C} \) to C).
02

Calculating Initial Potential Difference

Substitute the given values into the formula: \[ V = \frac{3.90 \times 10^{-6}}{920 \times 10^{-12}} \]Calculate \( V \) to find the potential difference between the plates.
03

Doubling Plate Separation

When the plate separation is doubled, the capacitance is halved because capacitance \( C \) is inversely proportional to the separation distance \( d \): \[ C \propto \frac{1}{d} \]Thus, the new capacitance \( C' = \frac{C}{2} = \frac{920 \times 10^{-12}}{2} \text{ F} \).
04

Calculating New Potential Difference

Using the new capacitance, apply the formula for potential difference again: \[ V' = \frac{Q}{C'} = \frac{3.90 \times 10^{-6}}{460 \times 10^{-12}} \]Calculate \( V' \), the new potential difference after doubling the separation.
05

Understanding Work Done in Doubling Separation

The work done \( W \) when changing the separation is related to the change in energy stored in the capacitor. The energy \( U \) stored in a capacitor is: \[ U = \frac{1}{2} C V^2 \]Calculate the initial energy \( U_i \) and the final energy \( U_f \), and find their difference: \[ W = U_f - U_i \]
06

Calculating Initial and Final Energy

Substitute the values for \( C \) and \( V \) into \( U = \frac{1}{2} C V^2 \) to get \( U_i \), and use \( C' \) and \( V' \) to get \( U_f \).Calculate \( W = U_f - U_i \).

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 Vaia!

Key Concepts

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

Parallel-Plate Capacitor
A parallel-plate capacitor is a device used to store electrical energy. Imagine it as two flat metal plates placed close to each other but not touching. These plates are separated by an insulating material called a dielectric, which, in this case, is air. The key function of these capacitors is to store electric charge when connected to a voltage source.

The ability of the capacitor to store charge depends on several factors:
  • The area of the plates: Larger plates can store more charge.
  • The distance between the plates: Closer plates increase capacitance.
  • The type of dielectric material: Different materials have varying capacities to store charge.

The relationship between the charge stored on the plates and the potential difference between them is defined by the capacitance value, which is constant for a given capacitor.
Potential Difference
Potential difference, often called voltage, is the force that pushes the electrical charge through the circuit. It's the difference in electric potential between two points. In a parallel-plate capacitor, this is the difference in electric potential between the two plates.

The formula for potential difference is derived from the charge and capacitance: \[ V = \frac{Q}{C} \]where:
  • \( V \) is the potential difference (in volts),
  • \( Q \) is the electrical charge (in coulombs),
  • \( C \) is the capacitance (in farads).

Calculating the potential difference involves plugging the charge and capacitance into the formula. This tells us how much energy each coulomb of charge has between the capacitor plates.
Electrical Charge
Electrical charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. It is measured in coulombs (C). In the scenario of a parallel-plate capacitor, the charge \( Q \) is built up on the plates as a result of the charging process, typically from a battery or power source.

The charge stored in a capacitor depends on the potential difference across its plates and its capacitance. When you increase the voltage or capacitance, more charge can be stored. Another important aspect is that the charge remains constant if the capacitor is isolated, meaning no additional energy source or conductor is connected to change it.
Energy Stored in Capacitor
Capacitors store energy in the electric field created between their plates. This stored energy can be used later when needed, which is why capacitors are crucial in electronic circuits. The amount of energy \( U \) stored is given by the formula:\[ U = \frac{1}{2} C V^2 \]where:
  • \( U \) is the energy stored (in joules),
  • \( C \) is the capacitance (in farads),
  • \( V \) is the potential difference (in volts).

This formula highlights that stored energy increases with both capacitance and the square of the voltage. Doubling the voltage quadruples the energy stored. Furthermore, if the capacitor's separation distance is altered, it changes the capacitance, affecting the energy storage and, ultimately, the work needed to separate the plates.

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

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00\(\times\) 10\(^6\) V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

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?

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\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

Your electronics company has several identical capacitors with capacitance \(C_1\) and several others with capacitance \(C_2\). You must determine the values of \(C_1\) and \(C_2\) but don't have access to \(C_1\) and \(C_2\) individually. Instead, you have a network with \(C_1\) and \(C_2\) connected in series and a network with \(C_1\) and \(C_2\) connected in parallel. You have a 200.0-V battery and instrumentation that measures the total energy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, 0.180 J of energy is stored in the network. When the series combination is connected, 0.0400 J of energy is stored. You are told that \(C_1\) is greater than \(C_2\). (a) Calculate \(C_1\) and \(C_2\). (b) For the series combination, does \(C_1\) or \(C_2\) store more charge, or are the values equal? Does \(C_1\) or \(C_2\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

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.

Sign-up for free