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Chapter 21: Problem 37

Two \(2.0 \mathrm{cm} \times 2.0 \mathrm{cm}\) square aluminum electrodes, spaced \(0.50 \mathrm{mm}\) apart, are connected to a \(100 \mathrm{V}\) battery. a. What is the capacitance? b. What is the charge on the positive electrode?

Short Answer

Expert verified
a. The capacitance is \(7.08 \times 10^{-12} \mathrm{F}\). b. The charge on the positive electrode is \(7.08 \times 10^{-10} \mathrm{C}\).

Step by step solution

01

Calculate the Area

First, convert the dimensions of the electrodes from cm to m for consistency in SI units. Therefore, \(2.0 \mathrm{cm}\) becomes \(0.02 \mathrm{m}\). Then calculate the area of one of the electrodes using the formula for the area of a square \(A = l \times l\), where \(l\) is the length of the side of the square. Here, \(A = (0.02 \mathrm{m})^2 = 0.0004 \mathrm{m}^2\).
02

Calculate the Capacitance

Utilize the formula for the capacitance of a parallel plate capacitor \(C = \epsilon_0 \times \frac{A}{d}\), where \(\epsilon_0\) is the permittivity of free space \((8.85 \times 10^{-12} \mathrm{F/m})\), \(A\) is the area of one of the plates \( (0.0004 \mathrm{m}^2)\), and \(d\) is the distance between the plates \(0.50 \mathrm{mm} = 0.00050 \mathrm{m}\). Calculate the capacitance \(C = 8.85 \times 10^{-12} \mathrm{F/m} \times \frac{0.0004 \mathrm{m}^2}{0.00050 \mathrm{m}} = 7.08 \times 10^{-12} \mathrm{F}\).
03

Calculate the Charge

Next, apply the relationship between charge, voltage and capacitance: \(Q = CV\). Use the calculated capacitance \( C = 7.08 \times 10^{-12} \mathrm{F} \) and given voltage \( V = 100 \mathrm{V} \) to find the charge: \(Q = 7.08 \times 10^{-12} \mathrm{F} \times 100 \mathrm{V} = 7.08 \times 10^{-10} \mathrm{C}\).

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Key Concepts

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

Electric Charge
Imagine electric charge as the fundamental property of matter that explains how objects attract or repel each other. It is quantified based on the presence of electrons or proton鈥攁 surplus or deficit of electrons results in a negative or positive charge, respectively. It is crucial for understanding electrical and magnetic phenomena. In the exercise provided, the concept of electric charge is explored through the calculation of the charge on the positive electrode of a parallel plate capacitor. This is done after finding the capacitor's capacitance and using the relationship given by the formula:
\( Q = CV \),
where \(Q\) represents the electric charge, \(C\) is the capacitance, and \(V\) denotes the electric potential (voltage).

To aid comprehension, consider how we use Coulombs (\(C\)) as the SI unit of electric charge, representing the amount of charge transported by a constant current of one ampere in one second.
Permittivity of Free Space
Permittivity of free space, denoted by \(\epsilon_0\), is a constant that represents how much electric field (\
SI Units
SI units, the International System of Units, provide a standard for measuring and communicating physical quantities, like length, mass, electric charge, and many others.

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Most popular questions from this chapter

Two 2.0 -cm-diameter disks spaced \(2.0 \mathrm{mm}\) apart form a parallel- plate capacitor. The electric field between the disks is \(5.0 \times 10^{5} \mathrm{V} / \mathrm{m}\) a. What is the voltage across the capacitor? b. How much charge is on each disk? c. An electron is launched from the negative plate. It strikes the positive plate at a speed of \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s} .\) What was the electron's speed as it left the negative plate?

A capacitor consists of two 6.0 -cm-diameter circular plates separated by \(1.0 \mathrm{mm} .\) The plates are charged to \(150 \mathrm{V},\) then the battery is removed. a. How much energy is stored in the capacitor? b. How much work must be done to pull the plates apart to where the distance between them is \(2.0 \mathrm{mm} ?\)

The capacity of a battery to deliver charge, and thus power, decreases with temperature. The same is not true of capacitors. For sure starts in cold weather, a truck has a 500 F capacitor alongside a battery. The capacitor is charged to the full \(13.8 \mathrm{V}\) of the truck's battery. How much energy does the capacitor store? How does the energy density of the \(9.0 \mathrm{kg}\) capacitor system compare to the \(130,000 \mathrm{J} / \mathrm{kg}\) of the truck's battery?

The highest magnetic fields in the world are generated when large arrays, or "banks," of capacitors are discharged through the copper coils of an electromagnet. At the National High Magnetic Field Laboratory, the total capacitance of the capacitor bank is 32 mF. These capacitors can be charged to \(16 \mathrm{kV}\). a. What is the energy stored in the capacitor bank when it is fully charged? b. When discharged, the entire energy from this bank flows through the magnet coil in 10 ms. What is the average power delivered to the coils during this time?

Raindrops acquire an electric charge as they fall. Suppose a \(2.5-\mathrm{mm}\) -diameter drop has a charge of \(+15 \mathrm{pC}\), fairly typical values. What is the potential at the surface of the raindrop?
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