Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 24: Problem 7

Which of the following is proportional to the capacitance of a parallel plate capacitor? a) the charge stored on each conducting plate b) the potential difference between the two plates c) the separation distance between the two plates d) the area of each plate e) all of the above f) none of the above

Short Answer

Expert verified
a) the charge stored on each conducting plate b) the potential difference between the two plates c) the separation distance between the two plates d) the area of each plate e) all of the above f) none of the above Answer: d) the area of each plate

Step by step solution

01

Recall the capacitance formula for a parallel plate capacitor

The formula for the capacitance (C) of a parallel plate capacitor is given as follows: C = \frac{k εₒ A}{d} Where εₒ (epsilon-naught) is the vacuum permittivity, k is dielectric constant, d is the separation distance between the two plates, and A is the area of each plate. By analyzing this formula, we can determine which of the given options is proportional to the capacitance.
02

Analyze the options according to the formula

Let's now analyze each option: a) the charge stored on each conducting plate: The capacitance formula does not have any term related to charges stored on plates. So, this option is not correct. b) the potential difference between the two plates: While potential difference does play a role in the overall formula, the relationship between it and the capacitance is indirect (Q=CV, where Q=charge and V=potential difference). The potential difference itself is not proportional to the capacitance. c) the separation distance between the two plates: The separation distance (d) is part of the formula but is inversely proportional to the capacitance (capacitance decreases as the distance between the plates increases). So, this option is not correct. d) the area of each plate: The area (A) of the plates is directly proportional to the capacitance (increased area leads to increased capacitance). So, this option is correct. e) all of the above: Based on the steps above, not all the options are correct. f) none of the above: Since option (d) is correct, this cannot be a valid choice. Conclusion: The correct answer to the question is option d) the area of each plate is proportional to the capacitance of a parallel plate capacitor.

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!

Key Concepts

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

Capacitance Formula
Understanding the capacitance formula for a parallel plate capacitor is essential in predicting how a capacitor will behave in an electrical circuit. The formula is expressed as

\[\begin{equation} C = \frac{k \varepsilon_0 A}{d} \end{equation}\]
where C represents capacitance, \( k \) is the dielectric constant, \( \varepsilon_0 \) is the electric permittivity of free space, A is the area of the capacitor's plates, and d is the separation distance between them. The direct proportionality between the capacitance and the plate area and dielectric constant indicates that increasing either will result in a higher capacitance, whereas increasing the separation distance will decrease the capacitance.
Dielectric Constant
The dielectric constant, often symbolized by \( k \), is a measure of a material's ability to store electrical energy within an electric field. When a dielectric material is inserted between the plates of a capacitor, it typically increases the capacitance in comparison to a vacuum (or air) between the plates.

Impact of Dielectric

Materials with a high dielectric constant are essentially better insulators and create capacitors with greater capacitance. This is an important concept when designing capacitors for specific applications, as the choice of dielectric material can significantly influence the efficiency and size of the capacitor.
Electrical Permittivity
Electrical permittivity, represented as \( \varepsilon \), refers to the measure of resistance that is encountered when forming an electric field within a particular medium. The vacuum permittivity, \( \varepsilon_0 \), is a fundamental physical constant and is approximately equal to

\[\begin{equation} 8.854 \times 10^{-12} \frac{\text{F}}{\text{m}} \end{equation}\]
(farads per meter). It essentially dictates how much electric field is 'permitted' to permeate through a vacuum. The higher the permittivity of a material in comparison to the vacuum permittivity, the greater the capacitance of a capacitor when the material is used as a dielectric.
Separation Distance
Separation distance between capacitor plates, noted as \( d \) in the capacitance formula, is inversely related to capacitance.

Inverse Proportionality

This means as the separation between the plates increases, the capacitance decreases. The physical explanation for this relationship is that a larger distance reduces the electric field strength between the plates for a given charge, in turn reducing the ability to store charge. Thus, capacitors with a small separation distance are generally preferred when a higher capacitance value is required.
Capacitor Plate Area
The plate area (A) of a parallel plate capacitor plays a pivotal role in determining its capacitance. The larger the area of the plates, the more charge they can store and, consequently, the greater the capacitance will be.

Increase in Capacitance

An increase in plate area results in more electric fields lines being able to terminate on the plate's surfaces, which corresponds to a higher ability to store electrical energy. This is why devices that require a large capacitance will often have capacitors with sizable plate areas.

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

Two parallel plate capacitors, \(C_{1}\) and \(C_{2},\) are connected in series to a \(96.0-\mathrm{V}\) battery. Both capacitors have plates with an area of \(1.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm} ;\) \(C_{1}\) has air between its plates, and \(C_{2}\) has that space filled with porcelain (dielectric constant of 7.0 and dielectric strength of \(5.70 \mathrm{kV} / \mathrm{mm}\) ). a) After charging, what are the charges on each capacitor? b) What is the total energy stored in the two capacitors? c) What is the electric field between the plates of \(C_{2} ?\)

A proton traveling along the \(x\) -axis at a speed of \(1.0 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters the gap between the plates of a \(2.0-\mathrm{cm}-\) wide parallel plate capacitor. The surface charge distributions on the plates are given by \(\sigma=\pm 1.0 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{2}\). How far has the proton been deflected sideways \((\Delta y)\) when it reaches the far edge of the capacitor? Assume that the electric field is uniform inside the capacitor and zero outside.

An \(8.00-\mu F\) capacitor is fully charged by a \(240 .-V\) battery, which is then disconnected. Next, the capacitor is connected to an initially uncharged capacitor of capacitance \(C,\) and the potential difference across it is found to be \(80.0 \mathrm{~V}\) What is \(C ?\) How much energy ends up being stored in the second capacitor?

A quantum mechanical device known as the Josephson junction consists of two overlapping layers of superconducting metal (for example, aluminum at \(1.00 \mathrm{~K}\) ) separated by \(20.0 \mathrm{nm}\) of aluminum oxide, which has a dielectric constant of \(9.1 .\) If this device has an area of \(100 . \mu \mathrm{m}^{2}\) and a parallel plate configuration, estimate its capacitance.

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.
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Fields in Physics

Read Explanation

Wave Optics

Read Explanation

Electricity and Magnetism

Read Explanation

Torque and Rotational Motion

Read Explanation

Circular Motion and Gravitation

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