/*! 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 6 Regarding the Earth and a cloud ... [FREE SOLUTION] | 91Ó°ÊÓ

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Regarding the Earth and a cloud layer \(800 \mathrm{m}\) above the Earth as the "plates" of a capacitor, calculate the capacitance. Assume the cloud layer has an area of \(1.00 \mathrm{km}^{2}\) and that the air between the cloud and the ground is pure and dry. Assume charge builds up on the cloud and on the ground until a uniform electric field of \(3.00 \times 10^{6} \mathrm{N} / \mathrm{C}\) throughout the space between them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold?

Short Answer

Expert verified
The maximum charge the cloud can hold is \(33.19 C\).

Step by step solution

01

Calculate the Capacity

The capacitance is given by the formula C = ε(A/d), where ε is the permittivity of free space, A is the area and d is the distance. In this scenario, the permittivity of free space \(ε\) is \(8.85 \times 10^{-12} \mathrm{m}^{-3} \mathrm{kg}^{-1} \mathrm{s}^{4} \mathrm{A}^{2}\). The distance \(d\) is the distance between the Earth and the cloud layer \(800m\). The area \(A\) is provided as \(1.00 \mathrm{km}^{2}\) which we need to convert to square meters so \(1.00 \mathrm{km}^{2} = 1.00 \times 10^{6} m^{2}\). So, the calculated capacity is \(C = (8.85 \times 10^{-12} \mathrm{m}^{-3} \mathrm{kg}^{-1} \mathrm{s}^{4} \mathrm{A}^{2}) (1.00 \times 10^{6} m^{2} / 800m) = 1.10625 \times 10^{-8} F\).
02

Calculate the Maximum Charge

Next, we need to calculate the maximum charge the cloud can hold using the formula Q = E * C. The electric field \(E\) is given as \(3.00 \times 10^{6} \mathrm{N} / \mathrm{C}\). Substituting the given value of \(E\) and the calculated capacitance \(C\) from step 1 into the formula, we get \(Q = (3.00 \times 10^{6} \mathrm{N} / \mathrm{C}) * (1.10625 \times 10^{-8} F) = 33.1875 C\). Therefore, 33.1875 Coulombs is the maximum charge the cloud can hold.

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Key Concepts

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

Understanding Electric Fields
An electric field describes the force that electric charges exert on each other. It is a vector field, meaning it has a direction as well as a magnitude, and is a cornerstone concept in electromagnetism. To visualize it, imagine a cloud with a buildup of electric charge hovering above the ground. This creates an invisible field in the space between the cloud and the ground, which, if you placed another charge within it, would experience a force. In our exercise, an electric field of \(3.00 \times 10^{6} \mathrm{N/C}\) is capable of ionizing air, allowing a current to pass through in the form of a lightning bolt. The strength of the field, measured in newtons per coulomb (N/C), directly influences how much charge can accumulate before the air 'breaks down' and conducts electricity.
Permittivity of Free Space
The permittivity of free space \(\varepsilon_0\), also called the electric constant, is a unit that appears in several key equations in electromagnetism. This fundamental constant represents the capability of the vacuum to permit electric field lines. Its value is approximately \(8.85 \times 10^{-12} \mathrm{m}^{-3} \mathrm{kg}^{-1} \mathrm{s}^{4} \mathrm{A}^{2}\). A material with high permittivity allows electric field lines to pass through it more easily and supports greater electric fields before breaking down. In our example, it's this constant that, when multiplied with the area of the cloud layer and divided by the distance to the ground, yields the capacitance of our Earth-cloud 'capacitor'.
Capacitance and Capacitors
A capacitor is essentially a device that stores electric potential energy by accumulating electric charge on two separated conductive plates. The measure of a capacitor's ability to store charge is called its capacitance, denoted by \(C\). Capacitance depends on the size of the plates (area \(A\)), the distance between them (\(d\)), and the permittivity of the material in between the plates (\(\varepsilon\)). The formula \(C = \frac{\varepsilon A}{d}\) demonstrates these relationships clearly. In this hypothetical exercise, the Earth and cloud are treated as the plates of a capacitor, with air as the dielectric material or 'separator'. Using the given values, we calculated the Earth-cloud capacitor's ability to hold a charge, which turns out to be 1.10625 \(\times\) 10^{-8} farads – a relatively small amount in terms of common capacitors but sufficient to understand the concept of capacitance related to natural phenomena such as lightning.

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Most popular questions from this chapter

To repair a power supply for a stereo amplifier, an electronics technician needs a \(100-\mu \mathrm{F}\) capacitor capable of withstanding a potential difference of \(90 \mathrm{V}\) between the plates. The only available supply is a box of five \(100-\mu \mathrm{F}\) capacitors, each having a maximum voltage capability of \(50 \mathrm{V} .\) Can the technician substitute a combination of these capacitors that has the proper electrical characteristics? If so, what will be the maximum voltage across any of the capacitors used? (Suggestion: The technician may not have to use all the capacitors in the box.)

An air-filled spherical capacitor is constructed with inner and outer shell radii of 7.00 and \(14.0 \mathrm{cm},\) respectively. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a charge of \(4.00 \mu \mathrm{C}\) on the capacitor?

A capacitor is constructed from two square plates of sides \(\ell\) and separation \(d\), as suggested in Figure \(\mathrm{P} 26.64 .\) You may assume that \(d\) is much less than \(\ell\). The plates carry charges \(+Q_{0}\) and \(-Q_{0} .\) A block of metal has a width \(\ell\), a length \(\ell\) and a thickness slightly less than \(d\). It is inserted a distance \(x\) into the capacitor. The charges on the plates are not disturbed as the block slides in. In a static situation, a metal prevents an electric field from penetrating inside it. The metal can be thought of as a perfect dielectric, with \(\kappa \rightarrow \infty,\) (a) Calculate the stored energy as a function of \(x\) (b) Find the direction and magnitude of the force that acts on the metallic block. (c) The area of the advancing front face of the block is essentially equal to \(\ell d\). Considering the force on the block as acting on this face, find the stress (force per area) on it. (d) For comparison, express the energy density in the electric field between the capacitor plates in terms of \(Q_{0}, \ell, d,\) and \(\epsilon_{0}\).

A 2.00 -nF parallel-plate capacitor is charged to an initial potential difference \(\Delta V_{i}=100 \mathrm{V}\) and then isolated. The dielectric material between the plates is mica, with a dielectric constant of \(5.00 .\) (a) How much work is required to withdraw the mica sheet? (b) What is the potential difference of the capacitor after the mica is withdrawn?

A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group?

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