/*! 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 9 A parallel plate capacitor is co... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 9

A parallel plate capacitor is connected to a battery for charging. After some time, while the battery is still connected to the capacitor, the distance between the capacitor plates is doubled. Which of the following is (are) true? a) The electric field between the plates is halved. b) The potential difference of the battery is halved. c) The capacitance doubles. d) The potential difference across the plates does not change. e) The charge on the plates does not change.

Short Answer

Expert verified
(Select all that apply.) a) The electric field between the plates is halved. b) The potential difference of the battery is halved. c) The capacitance doubles. d) The potential difference across the plates does not change. e) The charge on the plates does not change. Answer: a) The electric field between the plates is halved, d) The potential difference across the plates does not change, and e) The charge on the plates does not change are true statements.

Step by step solution

01

a) The electric field between the plates is halved

After doubling the distance, we can use the electric field formula to find out how the electric field changes: E_new = V/(2d), which is equal to (1/2) * V/d. This shows that the electric field (E_new) is indeed halved as compared to the initial electric field (E).
02

b) The potential difference of the battery is halved

The potential difference across the battery remains constant as it is not affected by the change in distance between the plates. Therefore, this statement is false.
03

c) The capacitance doubles

Using the capacitance formula for a parallel plate capacitor, we can see how the capacitance changes when the distance between the plates is doubled: C_new = 蔚鈧� * A/(2d). This is equal to (1/2) * 蔚鈧� * A/d which is half of the initial capacitance, not double. So, this statement is false.
04

d) The potential difference across the plates does not change

As the capacitor is still connected to the battery, the potential difference across the plates remains the same as it is equal to the potential difference of the battery. Therefore, this statement is true.
05

e) The charge on the plates does not change

As the potential difference remains the same and the charge formula is Q = CV, since we found out that C is halved, and V remains constant, the charge on the plates will also not change. Therefore, this statement is true. In conclusion, statements a), d), and e) are true, while statements b) and c) are false.

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 91影视!

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 capacitor of capacitance \(C\) has plates of area \(A\) with distance \(d\) between them. When the capacitor is connected to a battery of potential difference \(V\), it has a charge of magnitude \(Q\) on its plates. While the capacitor is connected to the battery, the distance between the plates is decreased by a factor of \(3 .\) The magnitude of the charge on the plates and the capacitance are then a) \(\frac{1}{3} Q\) and \(\frac{1}{3} C\). c) \(3 Q\) and \(3 C\). b) \(\frac{1}{3} Q\) and \(3 C\). d) \(3 Q\) and \(\frac{1}{3} C\).

The space between the plates of an isolated parallel plate capacitor is filled with a slab of dielectric material. The magnitude of the charge \(Q\) on each plate is kept constant. If the dielectric material is removed from between the plates, the energy stored in the capacitor a) increases. c) decreases. b) stays the same. d) may increase or decrease.

An isolated solid spherical conductor of radius \(5.00 \mathrm{~cm}\) is surrounded by dry air. It is given a charge and acquires potential \(V\), with the potential at infinity assumed to be zero. a) Calculate the maximum magnitude \(V\) can have. b) Explain clearly and concisely why there is a maximum.

A parallel plate capacitor consists of square plates of edge length \(2.00 \mathrm{~cm}\) separated by a distance of \(1.00 \mathrm{~mm}\). The capacitor is charged with a \(15.0-\mathrm{V}\) battery, and the battery is then removed. A \(1.00-\mathrm{mm}\) -thick sheet of nylon (dielectric constant \(=3.0\) ) is slid between the plates. What is the average force (magnitude and direction) on the nylon sheet as it is inserted into the capacitor?

A parallel plate capacitor with air in the gap between the plates is connected to a \(6.00-\mathrm{V}\) battery. After charging, the energy stored in the capacitor is \(72.0 \mathrm{~nJ}\). Without disconnecting the capacitor from the battery, a dielectric is inserted into the gap and an additional \(317 \mathrm{~nJ}\) of energy flows from the battery to the capacitor. a) What is the dielectric constant of the dielectric? b) If each of the plates has an area of \(50.0 \mathrm{~cm}^{2}\), what is the charge on the positive plate of the capacitor after the dielectric has been inserted? c) What is the magnitude of the electric field between the plates before the dielectric is inserted? d) What is the magnitude of the electric field between the plates after the dielectric is inserted?
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Dynamics

Read Explanation

Measurements

Read Explanation

Physics of Motion

Read Explanation

Circular Motion and Gravitation

Read Explanation

Electrostatics

Read Explanation
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.