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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.

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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.

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Most popular questions from this chapter

A parallel plate capacitor has a capacitance of \(120 .\) pF and a plate area of \(100 . \mathrm{cm}^{2}\). The space between the plates is filled with mica whose dielectric constant is \(5.40 .\) The plates of the capacitor are kept at \(50.0 \mathrm{~V}\) a) What is the strength of the electric field in the mica? b) What is the amount of free charge on the plates? c) What is the amount of charge induced on the mica?

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 parallel plate capacitor is constructed from two plates of different areas. If this capacitor is initially uncharged and then connected to a battery, how will the amount of charge on the big plate compare to the amount of charge on the small plate?

You have an electric device containing a \(10.0-\mu \mathrm{F}\) capacitor, but an application requires an \(18.0-\mu \mathrm{F}\) capacitor. What modification can you make to your device to increase its capacitance to \(18.0-\mu \mathrm{F} ?\)

When working on a piece of equipment, electricians and electronics technicians sometimes attach a grounding wire to the equipment even after turning the device off and unplugging it. Why would they do this?
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