/*! 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 3 A parallel-plate capacitor has c... [FREE SOLUTION] | 91Ó°ÊÓ

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A parallel-plate capacitor has circular plates of \(8.20 \mathrm{~cm}\) radius and \(1.30 \mathrm{~mm}\) separation. (a) Calculate the capacitance. (b) Find the charge for a potential difference of \(120 \mathrm{~V}\).

Short Answer

Expert verified
(a) Capacitance is 14.36 pF; (b) Charge is 1.723 nC.

Step by step solution

01

Convert Units

First, we need to convert all units to the standard SI units for consistency in calculations. The radius of the plates is given as \(8.20 \text{ cm}\), which converts to \(0.082 \text{ m}\). The separation between the plates is \(1.30 \text{ mm}\), which converts to \(0.0013 \text{ m}\).
02

Use the Capacitance Formula

The formula for the capacitance \(C\) of a parallel-plate capacitor is \(C = \varepsilon_0 \frac{A}{d}\), where \(A\) is the area of one plate and \(d\) is the separation between the plates. \(\varepsilon_0\) is the permittivity of free space, \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\).
03

Calculate Plate Area

The area \(A\) of a circular plate is given by \(A = \pi r^2\). Substituting the radius, \(A = \pi (0.082)^2 = 0.0211 \, \text{m}^2\).
04

Plug Values into Capacitance Formula

Now substitute the values into the capacitance formula: \[C = 8.85 \times 10^{-12} \frac{0.0211}{0.0013} \approx 1.436 \times 10^{-11} \, \text{F}\]
05

Capacitance in Picofarads

Convert the capacitance from Farads to picofarads: \(1.436 \times 10^{-11} \, \text{F}\) is equivalent to \(14.36 \, \text{pF}\) because \(1 \, \text{pF} = 10^{-12} \, \text{F}\).
06

Calculate the Charge

For part (b), use the formula \(Q = CV\), where \(Q\) is the charge and \(V\) is the potential difference. Substitute the capacitance and voltage: \[Q = 1.436 \times 10^{-11} \times 120 = 1.723 \times 10^{-9} \, \text{C}\]
07

Express Answer in Nanocoulombs

Convert the charge into nanocoulombs: \(1.723 \times 10^{-9} \, \text{C}\) is equivalent to \(1.723 \, \text{nC}\) because \(1 \, \text{nC} = 10^{-9} \, \text{C}\).

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Key Concepts

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

Capacitance
Capacitance is a measure of a capacitor's ability to store electric charge. It is the ratio of the amount of electric charge stored on the capacitor's plates to the potential difference between the plates. The standard unit for capacitance is the farad (F). In practice, capacitors often have microfarads (\( \mu \text{F} \)), nanofarads (\( \text{nF} \)), or picofarads (\( \text{pF} \)) due to the farad being a very large unit.

The formula to calculate the capacitance (\( C \)) of a parallel-plate capacitor is:\[ C = \varepsilon_0 \frac{A}{d} \]Where:
  • \( \varepsilon_0 \) is the permittivity of free space, a constant value of \( 8.85 \times 10^{-12} \, \text{F/m} \).
  • \( A \) represents the area of one of the capacitor's plates.
  • \( d \) is the separation distance between the plates.
In our example, converting the area from the radius and using the given separation, we calculated a capacitance of approximately \( 14.36 \, \text{pF} \).

This computation shows us how the capacitance is affected by both the size of the plates and the distance between them. Larger plate areas and smaller distances lead to higher capacitance.
Electric Charge
Electric charge is a fundamental property of matter. It can be positive or negative and is quantified in coulombs (C). Electrons are negatively charged, while protons carry a positive charge. In capacitors, electric charge is stored as an imbalance of electrons between two plates. This separation of charge creates an electric field.

For a given potential difference (\( V \)), the charge (\( Q \)) stored in a capacitor can be found using the formula:\[ Q = CV \]In this equation:
  • \( C \) is the capacitance.
  • \( V \) is the voltage applied across the capacitor.
Using this formula in our example, with a capacitance of \( 1.436 \times 10^{-11} \, \text{F} \) and a potential difference of \( 120 \, \text{V} \), we calculated the charge as \( 1.723 \times 10^{-9} \, \text{C} \).

The resulting charge transfer is equivalent to \( 1.723 \, \text{nC} \). This connection shows how the potential difference and capacitance determine the amount of charge a capacitor can hold.
Electric Potential
Electric potential, often referred to as voltage, is the potential energy per unit charge at a point in an electric field. It represents the capacity to do work in moving a charge from one place to another. The unit of electric potential is the volt (V).

In the context of capacitors, the electric potential difference, or simply the voltage, determines how much charge is stored in the capacitor. A higher potential difference means the capacitor can store more charge. The relationship between electric potential (\( V \)), charge (\( Q \)), and capacitance (\( C \)) is given by:
  • \( V = \frac{Q}{C} \)
  • or equivalently
    \( Q = CV \)
This formula unveils the linear relationship where a larger capacitance or higher voltage will lead to a greater charge being stored.

Thus, in our example, applying a potential of \( 120 \, \text{V} \) across the plates resulted in a calculated charge of \( 1.723 \, \text{nC} \), showing how critical the role of electric potential is in capacitor functionality.

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Most popular questions from this chapter

Two air-filled, parallel-plate capacitors are to be connected to a \(10 \mathrm{~V}\) battery, first individually, then in series, and then in parallel. In those arrangements, the energy stored in the capacitors turns out to be, listed least to greatest: \(75 \mu \mathrm{J}, 100 \mu \mathrm{J}, 300 \mu \mathrm{J}\), and \(400 \mu \mathrm{J} .\) Of the two capacitors, what is the (a) smaller and (b) greater capacitance?

The parallel plates in a capacitor, with a plate area of \(8.50 \mathrm{~cm}^{2}\) and an air-filled separation of \(3.00 \mathrm{~mm}\), are charged by a \(6.00 \mathrm{~V}\) battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of \(8.00 \mathrm{~mm}\). Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.

A coaxial cable used in a transmission line has an inner radius of \(0.10 \mathrm{~mm}\) and an outer radius of \(0.60 \mathrm{~mm}\). Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with polystyrene.

A parallel-plate capacitor has charge \(q\) and plate area \(A\). (a) By finding the work needed to increase the plate separation from \(x\) to \(x+d x\), determine the force between the plates. (Hint: See Eq.8-22.) (b) Then show that the force per unit area (the electrostatic stress) acting on either plate is equal to the energy density \(\varepsilon_{0} E^{2} / 2\) between the plates.

Two parallel plates of area \(100 \mathrm{~cm}^{2}\) are given charges of equal magnitudes \(8.9 \times 10^{-7} \mathrm{C}\) but opposite signs. The electric field within the dielectric material filling the space between the plates is \(1.4 \times 10^{6} \mathrm{~V} / \mathrm{m} .\) (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

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