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(1) Determine the capacitance of the Earth, assuming it to be a spherical conductor.

Short Answer

Expert verified
The Earth's capacitance is approximately \( 7.08 \times 10^{-4} \text{ F} \).

Step by step solution

01

Understanding the Formula for Capacitance

The formula for the capacitance \( C \) of a spherical conductor is given by: \[C = 4 \pi \varepsilon_0 R\]where \( R \) is the radius of the sphere and \( \varepsilon_0 \) is the permittivity of free space, approximately equal to \( 8.85 \times 10^{-12} \text{ F/m} \).
02

Identifying the Earth's Radius

The radius \( R \) of the Earth is approximately \( 6.371 \times 10^6 \text{ m} \). We will use this value to calculate the capacitance.
03

Plugging the Values into the Formula

Substitute \( R = 6.371 \times 10^6 \text{ m} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) into the formula \[C = 4 \pi \times 8.85 \times 10^{-12} \times 6.371 \times 10^6\]which simplifies to:
04

Calculating the Capacitance

Calculate the product \[C \approx 4 \times 3.14159 \times 8.85 \times 10^{-12} \times 6.371 \times 10^6 \approx 7.08 \times 10^{-4} \text{ F}\] Therefore, the capacitance of Earth is approximately \( 7.08 \times 10^{-4} \text{ F} \).

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Key Concepts

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

Spherical Conductor
A spherical conductor is essentially a sphere that can conduct electricity. This simple, symmetrical shape helps in easy calculation of certain electrical properties, such as capacitance. When we consider any spherical conductor, we assume it to be uniformly charged on its surface. This means that the electric field is directed radially outward and is the same at any point equidistant from the center. This characteristic plays a vital role in determining how the conductor interacts with its environment and in calculating electrical properties. An important feature of a spherical conductor is that its capacitance does not depend on its charge or potential but is solely a function of its size, represented by its radius, and the surrounding medium's properties. The simplicity of this characteristic makes spheres a popular model in physics problems to analyze fundamental electrical principles. For instance, we model the Earth as a spherical conductor to determine it's capacitance, offering a straightforward approach to understanding larger and more complex systems.
Permittivity of Free Space
The permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental physical constant essential in electromagnetism. It describes how an electric field affects, and is affected by, a vacuum. This constant is approximately \( 8.85 \times 10^{-12} \text{ F/m} \) (farads per meter). It's crucial for calculating the capacitance of conductors among other electrical properties.This permittivity value tells us how easily electric field lines can permeate the vacuum. In contexts where eventually there is no matter or air, such as between celestial bodies or in outer space, this constant comes into play. It's a key part of equations such as the formula for capacitance of a spherical conductor: \[C = 4 \pi \varepsilon_0 R\]where the capacitance \( C \) is directly proportional to \( \varepsilon_0 \). This means that any increase in permittivity would result in a higher capacitance, signifying a greater ability to store electrical charge.
Earth's Radius
Earth's radius is an important measurement in many scientific computations. For capacitance calculations, we use an average radius of approximately \( 6.371 \times 10^6 \text{ m} \). It's the distance from the center of the Earth to any point on its surface and is crucial when modeling the Earth as a spherical conductor.The uniformity of Earth’s radius assumption allows for simplification in calculations, giving us an effective way to apply geometric principles in physics. Naturally, Earth isn't a perfect sphere, but rather an oblate spheroid with slight bulging at the equator. However, for most practical purposes, especially in theoretical physics problems or global calculations, using the average radius suffices.This radius, when plugged into the capacitance formula along with the permittivity of free space, helps calculate the capacitance of the Earth, which is conceptually interesting as it represents the Earth's ability to store an electric charge. Understanding Earth's capacitance can also lead to better insights into how charge distribution occurs on larger, non-ideal bodies in space.

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Most popular questions from this chapter

(a) A gencral rule for estimating the capacitance \(C\) of an isolated conducting sphere with radius \(r\) is \(C\) (in \(\mathrm{pF}) \approx r(\) in \(\mathrm{cm})\) That is, the numerical value of \(C\) in \(\mathrm{pF}\) is about the same as the numerical value of the sphere's radius in \(\mathrm{cm}\). Justify this rule. (b) Modeling the human body as a \(1-\mathrm{m}\) -radius conducting sphere, use the given rule to estimate your body's capacitance. (c) While walking across a carpet, you acquire an excess "static electricity" charge \(Q\) and produce a \(0.5-\mathrm{cm}\) spark when reaching out to touch a metallic doorknob. The dielectric strength of air is \(30 \mathrm{kV} / \mathrm{cm}\). Use this information to estimate \(Q(\) in \(\mu \mathrm{C}) .\)

Two capacitors, \(C_{1}=3200 \mathrm{pF}\) and \(C_{2}=1800 \mathrm{pF},\) are connected in series to a 12.0-V battery. The capacitors are later disconnected from the battery and connected directly to each other, positive plate to positive plate, and negative plate to negative plate. What then will be the charge on each capacitor?

A parallel-plate capacitor is isolated with a charge \(\pm Q\) on each plate. If the separation of the plates is halved and a dielectric (constant \(K\) ) is inserted in place of air, by what factor does the energy storage change? To what do you attribute the change in stored potential energy? How does the new value of the electric field between the plates compare with the original value?

A \(2.70-\mu \mathrm{F}\) capacitor is charged to \(475 \mathrm{~V}\) and a \(4.00-\mu \mathrm{F}\) capacitor is charged to \(525 \mathrm{~V}\). \((a)\) These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor and the charge on each? (b) What is the voltage and charge for each capacitor if plates of opposite sign are connected?

A high-voltage supply can be constructed from a variable capacitor with interleaving plates which can be rotated as in Fig. \(24-36 .\) A version of this type of capacitor with more plates has a capacitance which can be varied from \(10 \mathrm{pF}\) to \(1 \mathrm{pF}\). (a) Initially, this capacitor is charged by a \(7500-\mathrm{V}\) power supply when the capacitance is \(8.0 \mathrm{pF}\). It is then disconnected from the power supply and the capacitance reduced to \(1.0 \mathrm{pF}\) by rotating the plates. What is the voltage across the capacitor now? (b) What is a major disadvantage of this as a high-voltage power supply?

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