/*! 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 15 Find the capacitance of the Eart... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the capacitance of the Earth. (Suggestion: The outer conductor of the "spherical capacitor" may be considered as a conducting sphere at infinity where \(V\) approaches zero.)

Short Answer

Expert verified
The capacitance of the Earth is approximately \(711 \, \mu F\).

Step by step solution

01

Identifying Known Values

Identify the known values from the problem. Here, we know that the radius of the Earth \(r = 6.37x10^6\) meters. The permittivity of free space \(\epsilon_o\) is a known constant and equals \(8.85x10^{-12}\) Farad/meter.
02

Apply Formula for Spherical Capacitance

Next, apply the formula for the capacitance of a spherical capacitor \(C = 4\pi \epsilon_o r\).
03

Solve for Capacitance

Substitute the known value of \(\epsilon_o\) and \(r\) into the formula \(C = 4\pi \epsilon_o r\) and calculate the value of \(C\), the capacitance of the Earth.

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Key Concepts

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

Spherical Capacitor
A spherical capacitor is a type of capacitor where the conductors are spherical in shape. It consists of two concentric spherical shells, one inside the other. However, when analyzing the capacitance of Earth, we consider a slightly different scenario: Earth itself acts as one conductor and a hypothetical spherical shell at infinity acts as the other.
In this situation, the outer shell being at "infinity" means it is far enough away so that we can assume the potential (\(V\)) there approaches zero. This reduces the complexity of the capacitor's geometry, allowing us to focus solely on the Earth's surface as a charged conductor.
Utilizing the concept of a spherical capacitor, we simplify calculations by assuming an infinite outer shell. This assumption is essential for understanding why formulas designed for capacitors with finite-sized conductors can adapt to work in this scenario.
Permittivity of Free Space
Permittivity of free space, symbolized by \( \epsilon_o \), is a fundamental physical constant crucial in electromagnetism. It characterizes how an electric field affects and is affected by a vacuum. Essentially, it measures the ability of a material (in this case, free space or vacuum) to permit electric field lines to pass through.
The value of permittivity of free space is approximately \( 8.85 \times 10^{-12} \) Farad per meter (F/m). This constant plays a vital role in calculating the capacitance of various objects, including Earth, by serving as a scaling factor in the capacitance formula.
In the context of a spherical capacitor or any capacitor, \( \epsilon_o \) relates the electric charge stored (or the potential difference applied) to the resulting capacitance. Recognizing this relationship enhances our comprehension of different electrical and electronic systems.
Capacitance Calculation
The capacitance of any object refers to its ability to store electrical charge. For a spherical body like Earth, the capacitance calculation is slightly different because a hypothetical spherical capacitor is modeled.
To calculate Earth's capacitance, we use the formula for a spherical capacitor: \( C = 4\pi \epsilon_o r \), where:
  • \( C \) is the capacitance.
  • \( \epsilon_o \) is the permittivity of free space.
  • \( r \) is the radius of Earth, in this case \( 6.37 \times 10^6 \) meters.
By substituting known values into this formula, we streamline the calculation process. The usage of constants like \( \epsilon_o \) simplifies the process, making it manageable to solve for capacitance even in complex configurations. This straightforward method allows us to determine how well Earth can store an electric charge in the context of its vast surface area and size.

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Most popular questions from this chapter

Show that the energy associated with a conducting sphere of radius \(R\) and charge \(Q\) surrounded by a vacuum is \(U=k_{e} Q^{2} / 2 R\).

Example 26.2 explored a cylindrical capacitor of length \(\ell\) and radii \(a\) and \(b\) of the two conductors. In the What If? section, it was claimed that increasing \(\ell\) by \(10 \%\) is more effective in terms of increasing the capacitance than increasing \(a\) by \(10 \%\) if \(b>2.85 a\). Verify this claim mathematically.

A parallel-plate capacitor has a charge \(Q\) and plates of area \(A\). What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is \(E=Q / A \epsilon_{0},\) you might think that the force is \(F=Q E=Q^{2} / A \epsilon_{0} .\) This is wrong, because the field \(E\) includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually \(F=Q^{2} / 2 \epsilon_{0} A .\) (Suggestion: Let \(C=\epsilon_{0} A / x\) for an arbitrary plate separation \(x ;\) then require that the work done in separating the two charged plates be \(W=\int F \, d x\).) The force exerted by one charged plate on another is sometimes used in a machine shop to hold a workpiece stationary.

Two capacitors, \(C_{1}=5.00 \mu \mathrm{F}\) and \(C_{2}=12.0 \mu \mathrm{F},\) are connected in parallel, and the resulting combination is connected to a \(9.00-\mathrm{V}\) battery. (a) What is the equivalent capacitance of the combination? What are (b) the potential difference across each capacitor and (c) the charge stored on each capacitor?

Review problem. A certain storm cloud has a potential of \(1.00 \times 10^{8} \mathrm{V}\) relative to a tree. If, during a lightning storm, \(50.0 \mathrm{C}\) of charge is transferred through this potential difference and \(1.00 \%\) of the energy is absorbed by the tree, how much sap in the tree can be boiled away? Model the sap as water initially at \(30.0^{\circ} \mathrm{C}\). Water has a specific heat of \(4186 \mathrm{J} / \mathrm{kg}^{\circ} \mathrm{C},\) a boiling point of \(100^{\circ} \mathrm{C},\) and a latent heat of vaporization of \(2.26 \times 10^{6} \mathrm{J} / \mathrm{kg}\).

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