/*! 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 46 A parallel-plate capacitor \(C\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A parallel-plate capacitor \(C\) is charged up to a potential \(V_{0}\) with a charge of magnitude \(Q_{0}\) on each plate. It is then disconnected from the battery, and the plates are pulled apart to twice their original separation. (a) What is the new capacitance in terms of \(C ?\) (b) How much charge is now on the plates in terms of \(Q_{0} ?(\mathrm{c})\) What is the potential difference across the plates in terms of \(V_{0} ?\)

Short Answer

Expert verified
(a) \(\frac{C}{2}\); (b) \(Q_0\); (c) \(2V_0\).

Step by step solution

01

Understand the Initial Setup of the Capacitor

Initially, we have a parallel-plate capacitor with capacitance \(C\), charged to a potential \(V_0\), and with a charge \(Q_0\). The capacitance \(C\) is given by \(C = \frac{\varepsilon_0 A}{d}\), where \(A\) is the plate area, \(d\) is the separation between the plates, and \(\varepsilon_0\) is the permittivity of free space. The charge \(Q_0\) and the potential \(V_0\) relate to each other by \(Q_0 = C \cdot V_0\).
02

Calculate the New Capacitance

The plates are pulled apart to twice their original separation, so the new separation \(d' = 2d\). The capacitance of a parallel-plate capacitor is inversely proportional to the separation distance: \(C' = \frac{\varepsilon_0 A}{d'} = \frac{\varepsilon_0 A}{2d} = \frac{C}{2}\). Thus, the new capacitance \(C'\) is \(\frac{C}{2}\).
03

Determine the Charge After Disconnection

Since the battery is disconnected before the plates are moved, charge cannot enter or leave the plates. Thus, the charge remains the same: \(Q' = Q_0\).
04

Calculate the New Potential Difference

Using the relation \(V' = \frac{Q'}{C'}\) where \(Q' = Q_0\) and \(C' = \frac{C}{2}\), we find the new potential \(V' = \frac{Q_0}{\frac{C}{2}} = \frac{2Q_0}{C} = 2V_0\). The new potential difference across the plates is \(2V_0\).

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 is a device that stores electrical energy by maintaining a separation of charges on two conductive plates. These plates are arranged with an insulating material, or dielectric, sandwiched between them. The basic setup involves two plates that are parallel to each other, which helps in achieving a uniform electric field between the plates. In this setup, the distance between the plates and the surface area of the plates are key factors that affect the storing capacity of the capacitor. By changing the distance between these plates, we can influence the capacitance, as seen in the problem above when the plates are pulled apart. The uniformity of the electric field makes parallel-plate capacitors popular in various electronic applications where predictable behavior is needed.
Capacitance
Capacitance is a measure of a capacitor's ability to store charge. It is defined by the capacitance equation: \[C = \frac{\varepsilon_0 A}{d}\] where:
  • \(C\) is the capacitance,
  • \(\varepsilon_0\) represents the permittivity of free space,
  • \(A\) is the area of one of the plates,
  • \(d\) is the separation between the plates.
Capacitance is directly proportional to the plate area and inversely proportional to the distance between the plates. This means if you increase the area while keeping everything else constant, the capacitance increases. However, if the separation distance is doubled, the capacitance is halved, as seen when the plates are pulled apart to twice their original separation, causing the capacitance to change from \(C\) to \(\frac{C}{2}\). This inversely proportional relationship between capacitance and distance is a critical factor in the design and application of capacitors.
Electric Potential Difference
The electric potential difference, also known as voltage, between two points in a circuit is a measure of the energy difference per charge between those points. For a capacitor, this is the voltage across its plates. Initially, the potential difference across the plates is given by \(V_0\). When the distance between the plates is altered, the capacitance changes, affecting the potential difference. After the battery is disconnected and the plates are pulled apart, the potential difference increases. The change in potential difference can be calculated using the formula:\[V' = \frac{Q'}{C'}\]With the charge remaining constant after disconnection \((Q' = Q_0)\) and the capacitance halved \((C' = \frac{C}{2})\), the new potential difference becomes twice the original, \(2V_0\). This illustrates how potential difference adjusts to maintain the relationship with charge and capacitance.
Charge Conservation
Charge conservation is a fundamental principle that states the total electric charge in an isolated system remains constant over time. This principle applies to the parallel-plate capacitor in our scenario. When the capacitor is disconnected from the battery, it becomes an isolated system with respect to its charge. Therefore, even though the plates are physically moved, resulting in a change in capacitance and potential difference, the charge \(Q\), originally \(Q_0\), remains unchanged. This demonstrates that while capacitors are capable of changing how charge is distributed or stored depending on physical alterations to the system, they strictly adhere to the law of conservation of charge, which is crucial for maintaining balance in electrical circuits.

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) How much charge does a battery have to supply to a 5.0\(\mu \mathrm{F}\) capacitor to create a potential difference of 1.5 \(\mathrm{V}\) across its plates? How much energy is stored in the capacitor in this case? (b) How much charge would the battery have to supply to store 1.0 \(\mathrm{J}\) of energy in the capacitor? What would be the potential across the capacitor in that case?

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 \(1.00 \times 10^{2} \mathrm{V}\) . The electric field between he plates is to be no greater than \(1.00 \times 10^{4} \mathrm{N} / \mathrm{C} .\) 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 and (b) find the maximum charge these plates can hold.

In a particular Millikan oil-drop apparatus, the plates are 2.25 \(\mathrm{cm}\) apart. The oil used has a density of \(0.820 \mathrm{g} / \mathrm{cm}^{3},\) and the atomizer that sprays the oil drops produces drops of diameter \(1.00 \times 10^{-3} \mathrm{mm}\) . (a) What strength of electric field is needed to hold such a drop stationary against gravity if the drop contains five excess electrons? (b) What should be the potential difference across the plates to produce this electric field? (c) If another drop of the same oil requires a plate potential of 73.8 \(\mathrm{V}\) to hold it stationary, how many excess electrons did it contain?

A point charge has a charge of \(2.50 \times 10^{-11} \mathrm{C}\). At what distance from the point charge is the electric potential (a) \(90.0 \mathrm{~V} ?\) (b) \(30.0 \mathrm{~V}\) ? Take the potential to be zero at an infinite distance from the charge.

How far from a \(-7.20 \mu \mathrm{C}\) point charge must a \(+2.30 \mu \mathrm{C}\) point charge be placed in order for the electric potential energy of the pair of charges to be \(-0.400 \mathrm{J} ?\) (Take the energy to be zero when the charges are infinitely far apart.)

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.