/*! 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 103 A parallel-plate capacitor is co... [FREE SOLUTION] | 91Ó°ÊÓ

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A parallel-plate capacitor is connected to a battery that maintains a constant potential difference between the plates. If the spacing between the plates is doubled, does the magnitude of the charge on the plates increase, decrease, or stay the same?

Short Answer

Expert verified
The charge on the plates decreases.

Step by step solution

01

Identify Initial Conditions

The capacitor is connected to a battery that keeps the potential difference constant. This setup ensures that the voltage across the capacitor remains unchanged, regardless of modifications to the capacitor's parameters.
02

Recall Capacitance Formula

The formula for the capacitance of a parallel-plate capacitor is given by: \[ C = \frac{\varepsilon_0 A}{d} \]where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of one of the plates, and \( d \) is the separation between the plates.
03

Relate Capacitance to Charge

The charge \( Q \) on a capacitor is related to its capacitance \( C \) and the potential difference \( V \) by the formula: \[ Q = C \times V \]Since \( V \) is constant (due to the battery), any change in capacitance \( C \) will directly affect the charge \( Q \).
04

Evaluate the Effect of Doubling the Distance

Doubling the spacing \( d \) between the plates leads to a new capacitance: \[ C' = \frac{\varepsilon_0 A}{2d} = \frac{1}{2} \times \frac{\varepsilon_0 A}{d} = \frac{C}{2} \]The capacitance is halved when the distance is doubled.
05

Determine the New Charge

Since \( V \) is constant and the new capacitance \( C' = \frac{C}{2} \), the new charge \( Q' \) becomes:\[ Q' = C' \times V = \frac{C}{2} \times V = \frac{Q}{2} \]The magnitude of the charge on the plates is reduced to half its original value.

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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 simple device consisting of two conductive plates separated by a small distance. It is widely used in electronic circuits to store electric charge temporarily. The key focus of a parallel-plate capacitor is its ability to maintain an electric field between its plates when a voltage, or potential difference, is applied. This is achieved because the plates are connected to a power source like a battery.
As a result, an electric field is established between the plates, allowing the capacitor to store energy. This stored energy becomes useful in various applications, such as filtering noise in circuits or smoothing capacitive loads in power supplies. However, the effectiveness of a parallel-plate capacitor depends significantly on several parameters including the area of the plates, the distance between them, and the dielectric material separating the plates.
  • The area of the plates: Larger plate areas allow for the storage of more charge.
  • The distance between the plates: Smaller distances increase the capacitance and the ability to store charge.
  • The dielectric material: Certain materials can enhance the capacitor's ability to store charge by increasing its capacitance.
Capacitance
Capacitance is a fundamental concept that describes how much charge a capacitor can hold for a given voltage. For a parallel-plate capacitor, the capacitance formula is: \[ C = \frac{\varepsilon_0 A}{d} \]where:
  • \( C \) is the capacitance, measured in farads (F).
  • \( \varepsilon_0 \) is the permittivity of free space, a constant that affects how an electric field interacts with space.
  • \( A \) is the area of one of the plates, measured in square meters.
  • \( d \) is the distance between the plates, measured in meters.
This formula shows that capacitance is directly proportional to the plate area and inversely proportional to the distance between the plates. Consequently, when the distance \( d \) between the plates is doubled, as in the given exercise, the capacitance is halved. This reduction in capacitance means the capacitor can hold less charge at the same voltage. In practice, knowing how these factors interplay helps in designing capacitors that meet specific needs in electronic circuits. By choosing the right materials and physical dimensions, engineers can shape the capacitor's characteristics to best fit their application's requirements.
Electric Charge
Electric charge is a property of matter that causes it to experience a force when placed in an electric field. In capacitors, electric charge is what gets stored and released, making capacitors essential components for managing charge in electrical circuits. The amount of charge \( Q \) stored in a capacitor is related to its capacitance \( C \) and the voltage \( V \) across it through the equation:\[ Q = C \times V \]Here, if the voltage \( V \) is kept constant, as is the case with a battery connected to the capacitor, any change in capacitance \( C \) will directly alter the charge \( Q \). If you halve the capacitance, as seen in the exercise when the plate separation is doubled, the charge stored will also be halved. This simple relationship underscores the importance of understanding both capacitance and electric charge when working with capacitors. Additionally, this principle allows more complex aspects of circuits, such as timing and filtering, to be effectively controlled and manipulated.

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Most popular questions from this chapter

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a \(0.250-\mathrm{kg}\) ball with zero net charge was dropped from rest from a height of \(1.00 \mathrm{~m}\). The ball landed \(0.552 \mathrm{~s}\) later. Next, the ball was given a net charge of \(7.75 \mu \mathrm{C}\) and dropped from rest from the same height. This time the ball fell for \(0.680 \mathrm{~s}\) before landing. What is the electric potential at a height of \(1.00 \mathrm{~m}\) above the ground on this planet, given that the electric potential at ground level is zero? (Air resistance can be ignored.)

A capacitor stores \(2.7 \times 10^{-5} \mathrm{C}\) of charge when it is connected to a \(9.0-\mathrm{V}\) battery. What is the capacitance of the capacitor?

An electron accelerates from rest through a region where the electric potential increases by the amount \(V\). If the final speed of the electron is \(4.8 \times 10^{4} \mathrm{~m} / \mathrm{s}\), what is the potential difference?

An electric field does \(0.052 \mathrm{~J}\) of work as a \(+5.7-\mu \mathrm{C}\) charge moves from point A to point B. Find the difference in electric potential, \(\Delta V=V_{\mathrm{B}}-V_{\mathrm{A}}\), between the points \(\mathrm{A}\) and \(\mathrm{B}\).

A capacitor with a capacitance of \(430 \mathrm{pF}\) is connected to a battery with a voltage of \(550 \mathrm{~V}\). (a) What is the magnitude of the charge on each plate of the capacitor? (b) How much energy is stored in the capacitor? (c) What is the electric field between the plates if their separation is \(0.89 \mathrm{~mm}\) ?

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