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Chapter 18: Problem 83

A capacitor consists of two parallel plates, each with an area of 16.0 \(\mathrm{cm}^{2}\) , separated by a distance of 0.200 \(\mathrm{cm} .\) The material that fills the volume between the plates has a dielectric constant of \(5.00 .\) The plates of the capacitor are connected to a \(300-\mathrm{V}\) battery. (a) What is the capacitance of the capacitor? (b) What is the charge on either plate? (c) How much energy is stored in the charged capacitor?

Short Answer

Expert verified
Capacitance = 35.4 pF; Charge = 10.6 nC; Energy = 1.59 渭J.

Step by step solution

01

Convert Units to SI

First, convert the given area and separation distance to \(\mathrm{m}^2\) and meters, respectively, because the SI units for area is \(\mathrm{m}^2\) and for distance is meters. - Area \(A = 16.0\ \mathrm{cm}^2 = 16.0 \times 10^{-4} \ \mathrm{m}^2\) - Separation distance \(d = 0.200\ \mathrm{cm} = 0.00200\ \mathrm{m}\).
02

Calculate the Capacitance

Use the formula for the capacitance of a parallel-plate capacitor filled with a dielectric: \(C = \varepsilon_r \varepsilon_0 \frac{A}{d}\)where \(\varepsilon_r = 5.00\) is the dielectric constant, \(\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{F/m}\) is the permittivity of free space, \(A\) is the area of the plates, and \(d\) is the separation distance.Substitute the given values: \C = 5.00 \times 8.85 \times 10^{-12} \frac{16.0 \times 10^{-4}}{0.00200} \C = 3.54 \times 10^{-11} \, \mathrm{F} = 35.4\, \mathrm{pF}\.
03

Calculate the Charge on Either Plate

The charge \(Q\) on either plate can be found using the formula: \Q = CV\where \(V\) is the voltage provided by the battery, \(V = 300\, \mathrm{V}\), and \(C\) is the capacitance from Step 2.Substitute the known values: \Q = 3.54 \times 10^{-11}\, \mathrm{F} \times 300\, \mathrm{V}\Q = 1.06 \times 10^{-8}\, \mathrm{C}.\.
04

Calculate the Energy Stored in the Capacitor

Calculate the energy \(U\) stored in the capacitor using the formula: \U = \frac{1}{2}CV^2\Substitute the value of capacitance from Step 2 and voltage: \U = \frac{1}{2} \times 3.54 \times 10^{-11}\, \mathrm{F} \times (300\, \mathrm{V})^2\U = 1.59 \times 10^{-6}\, \mathrm{J}\.

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Key Concepts

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

Dielectric Constant
The dielectric constant is a crucial factor in determining the capacitance of a capacitor. It is a measure of the material's ability to increase the storage capacity of electric charge when placed between the plates of a capacitor. This constant is denoted as \(\varepsilon_r\) and is dimensionless.
For instance, in a capacitor with a dielectric material, the capacitance is not just governed by the physical characteristics like plate area and separation distance, but also by this constant.
  • A higher dielectric constant means the material can store more charge under the same electric field.
  • It effectively increases the capacitance compared to air or vacuum-filled capacitors.
In this exercise, the dielectric constant value of 5.00 significantly boosts the capacitor's ability to store electrical energy.
Energy Storage in Capacitors
Energy storage in capacitors is an important concept as it involves the ability of these devices to hold electrical energy in an electric field. The energy stored (U) in a capacitor is determined by its capacitance (C) and the voltage (V) across it.
This relationship is given by the formula: \[U = \frac{1}{2}CV^2\]
  • Capacitance determines how much energy the capacitor can store for a given voltage.
  • This ability to store energy makes capacitors very useful in a variety of electrical and electronic applications.
In the provided exercise, we calculated the energy stored as 1.59 \times 10^{-6} \mathrm{J}, demonstrating the capacitor's capability given the conditions provided.
Parallel-Plate Capacitor
A parallel-plate capacitor is one of the simplest types of capacitors and consists of two conductive plates separated by an insulating material or a dielectric. This setup is critical because it enhances the capacitor's ability to store electric charge.
The formula for calculating the capacitance (C) of this type of capacitor is:\[C = \varepsilon_r \varepsilon_0 \frac{A}{d}\]
  • \(\varepsilon_0\) is the permittivity of free space, an inherent property of the vacuum.
  • A is the area of each plate.
  • d is the separation distance between the plates.
In the exercise, the use of a dielectric increases the effectiveness of this configuration, heightening the capacitance to \(35.4\, \mathrm{pF}\).
SI Units Conversion
SI units conversion is vital in physics to ensure uniformity and precision in calculations. It involves changing units to the standard International System of Units (SI) to maintain consistency across scientific disciplines.
In the initial step of the exercise, we converted units to meter-squared (m虏) for area and meters (m) for distance.
  • Infractions like \(16.0\, \mathrm{cm}^2\) become \(16.0 \times 10^{-4} \, \mathrm{m}^2\).
  • Distances like \(0.200\, \mathrm{cm}\) are equivalent to \(0.00200\, \mathrm{m}\).
These conversions are crucial for using standard formulas in physics, allowing for accurate comparative analysis and universal understanding of measurements.

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Most popular questions from this chapter

(a) If an electron and a proton each have a kinetic energy of 1.00 eV, how fast is each one moving? (b) What would be their speeds if each had a kinetic energy of 1.00 \(\mathrm{keV}\) ? (c) If they were each traveling at 1.00\(\%\) the speed of light, what would be their kinetic energies in keV?

A parallel-plate capacitor is to be constructed by using, as a dielectric, rubber with a dielectric constant of 3.20 and a dielectric strength of 20.0 \(\mathrm{MV} / \mathrm{m}\) . The capacitor is to have a capacitance of 1.50 \(\mathrm{nF}\) and must be able to withstand a maximum potential difference of 4.00 \(\mathrm{kV} .\) What is the minimum area the plates of this capacitor can have?

A point charge \(Q=+4.60 \mu \mathrm{C}\) is held fixed at the origin. A second point charge \(q=+1.20 \mu \mathrm{C}\) with mass of \(2.80 \times\) \(10^{-4} \mathrm{kg}\) is placed on the \(x\) axis, 0.250 \(\mathrm{m}\) from the origin. (a) What is the electric potential energy \(U\) of the pair of charges? (Take \(U\) to be zero when the charges have infinite separation.) (b) The second point charge is released from rest. What is its speed when its distance from the origin is (i) \(0.500 \mathrm{m} ;\) (ii) 5.00 \(\mathrm{m}\) ; (iii) 50.0 \(\mathrm{m} ?\)

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{\boldsymbol{E}}\) directed to the left. It is released from rest and moves to the left; after it has moved \(6.00 \mathrm{cm},\) its kinetic energy is found to be \(+1.50 \times 10^{-6} \mathrm{J}\) (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the endpoint? (c) What is the magnitude of \(\vec{E} ?\)

A 10.0\(\mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a 12.0 \(\mathrm{V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0 \(\mathrm{V}\) battery after the radius of each plate was doubled without changing their separation?
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