Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 24: Problem 24

A parallel plate capacitor is connected to a battery. As the plates are moved farther apart, what happens to each of the following? a) the potential difference across the plates b) the charge on the plates c) the electric field between the plates

Short Answer

Expert verified
Answer: As the distance between the plates increases, the potential difference across the plates remains the same, the charge on the plates remains the same, and the electric field between the plates decreases.

Step by step solution

01

1. Recall Capacitance Formula

To start, let's recall the capacitance formula for a parallel plate capacitor, which is given by: C = ε₀ * A / d, where C is the capacitance, ε₀ is the vacuum permittivity, A is the area of the plates, and d is the distance between the plates.
02

2. Analyze the Potential Difference

The potential difference across the capacitor can be found using the formula: V = Q / C, where V is the voltage, Q is the charge on the plates, and C is the capacitance. As the distance between the plates (d) increases, the capacitance (C) decreases based on the capacitance formula. Since the capacitor is connected to a battery, the potential difference remains constant. Therefore, the potential difference across the plates does not change.
03

3. Analyze the Charge on the Plates

As the potential difference across the plates remains constant due to the battery, and the capacitance decreases as the plates are moved farther apart, the charge on the plates (Q) can be calculated as: Q = C * V. Since V is constant, as the capacitance decreases, the charge on the plates remains the same.
04

4. Analyze the Electric Field between the Plates

The electric field between the plates can be calculated using the formula: E = V / d, where E is the electric field, V is the voltage, and d is the distance between the plates. As the plates are moved farther apart (d increases), the potential difference (V) remains constant, so the electric field between the plates decreases as the distance increases. In summary: a) The potential difference across the plates remains the same. b) The charge on the plates remains the same. c) The electric field between the plates decreases.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The Earth is held together by its own gravity. But it is also a charge-bearing conductor. a) The Earth can be regarded as a conducting sphere of radius \(6371 \mathrm{~km},\) with electric field \(\vec{E}=(-150 . \mathrm{V} / \mathrm{m}) \hat{r}\) at its surface, where \(\hat{r}\) is a unit vector directed radially outward. Calculate the total electrostatic potential energy associated with the Earth's electric charge and field. b) The Earth has gravitational potential energy, akin to the electrostatic potential energy. Calculate this energy, treating the Earth as a uniform solid sphere. (Hint: \(d U=-(G m / r) d m\). c) Use the results of parts (a) and (b) to address this question: To what extent do electrostatic forces affect the structure of the Earth?

A dielectric slab with thickness \(d\) and dielectric constant \(\kappa=2.31\) is inserted in a parallel place capacitor that has been charged by a \(110 .-\mathrm{V}\) battery and having area \(A=\) \(100, \mathrm{~cm}^{2},\) and separation distance \(d=2.50 \mathrm{~cm}\). a) Find the capacitance, \(C,\) the potential difference, \(V,\) the electric field, \(E,\) the total charge stored on the capacitor \(Q\), and electric potential energy stored in the capacitor, \(U\), before the dielectric material is inserted. b) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab has been inserted and the battery is still connected. c) Find \(C, V, E, Q\) and \(U\) when the dielectric slab is in place and the battery is disconnected.

A capacitor with a vacuum between its plates is connected to a battery and then the gap is filled with Mylar. By what percentage is its energy-storing capacity increased?

Consider a cylindrical capacitor, with outer radius \(R\) and cylinder separation \(d\). Determine what the capacitance approaches in the limit where \(d \ll R\). (Hint: Express the capacitance in terms of the ratio \(d / R\) and then examine what happens as the ratio \(d / R\) becomes very small compared to \(1 .)\) Explain why the limit on the capacitance makes sense.

The capacitance of a spherical capacitor consisting of two concentric conducting spheres with radii \(r_{1}\) and \(r_{2}\) \(\left(r_{2}>r_{1}\right)\) is given by \(C=4 \pi \epsilon_{0} r_{1} r_{2} /\left(r_{2}-r_{1}\right) .\) Suppose that the space between the spheres, from \(r,\) up to a radius \(R\) \(\left(r_{1}
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Astrophysics

Read Explanation

Particle Model of Matter

Read Explanation

Fundamentals of Physics

Read Explanation

Engineering Physics

Read Explanation

Physical Quantities And Units

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.

Sign-up for free