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The Earth can be thought of as a spherical capacitor. If the net charge on the Earth is \(-7.8 \cdot 10^{5} \mathrm{C},\) find \((\) a) the capacitance of the Earth and (b) the electric potential energy stored on the Earth's surface

Short Answer

Expert verified
Answer: The capacitance of the Earth as a spherical capacitor is \(7.154 \cdot 10^{-5} \mathrm{F}\) and the electric potential energy stored on its surface is \(2.673 \cdot 10^{11} \mathrm{J}\).

Step by step solution

01

Identify the given values

We are given the net charge on the Earth \(Q = -7.8 \cdot 10^{5} \mathrm{C}\) and we know that the radius of the Earth \(R = 6.37 \cdot 10^{6}\,\text{m}\). We also need the value of the vacuum permittivity constant, which is \(\epsilon_{0} = 8.854 \cdot 10^{-12}\,\text{F/m}\).
02

Find the capacitance of the Earth

To find the capacitance of the Earth as a spherical capacitor, we use the formula for capacitance of a sphere: \(C = 4\pi \epsilon_0 R\). Plugging in the values, we get: \(C = 4\pi \cdot (8.854 \cdot 10^{-12}\,\mathrm{F/m}) \cdot (6.37 \cdot 10^{6}\,\mathrm{m})\) \(C = 7.154 \cdot 10^{-5} \mathrm{F}\) So, the capacitance of the Earth is \(7.154 \cdot 10^{-5} \mathrm{F}\).
03

Find the electric potential energy stored on the Earth's surface

To find the electric potential energy stored on the Earth's surface, we use the formula for the energy stored in a capacitor: \(U = \frac{1}{2} C V^2\). We know that \(V = \frac{Q}{C}\), so we can rewrite the formula as: \(U = \frac{1}{2} C (\frac{Q}{C})^2\) Now, we plug in the values for \(Q\) and \(C\): \(U = \frac{1}{2} \cdot (7.154 \cdot 10^{-5} \mathrm{F}) \cdot (\frac{-7.8 \cdot 10^{5} \mathrm{C}}{7.154 \cdot 10^{-5} \mathrm{F}})^2\) \(U = 2.673 \cdot 10^{11} \mathrm{J}\) The electric potential energy stored on the Earth's surface is \(2.673 \cdot 10^{11} \mathrm{J}\).

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Key Concepts

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

Earth's Capacitance
The concept of Earth's capacitance emerges from the idea of treating our planet as a gigantic spherical capacitor. It's a fascinating application of electrostatic principles on a planetary scale, which can help us understand the electrical properties of Earth. The capacitance of an object indicates its ability to store electrical charge, and when we speak of Earth's capacitance, we are referring to its potential to hold an electrical charge against the vacuum of space around it.

When we apply this idea to Earth, we consider Earth as a conducting sphere and space as the other conductor with an infinitely large radius. The capacitance formula for a sphere is given by the expression \(C = 4\pi \epsilon_{0} R\), where \(C\) is the capacitance, \(\epsilon_{0}\) represents the vacuum permittivity or the electric constant, and \(R\) is the radius of the sphere. To enhance student comprehension, imagine that Earth's charge is evenly distributed on its surface, and the vacuum around it serves as the dielectric material.

By knowing Earth's radius and the constant \(\epsilon_{0}\), one can calculate Earth's ability to store charge. Understanding the Earth's capacitance not only highlights the scale of charges involved in global phenomena but also offers insight into how Earth might interact with its space environment, including solar winds and atmospheric electricity.
Electric Potential Energy
Electric potential energy in the context of capacitors is a crucial concept that defines the energy a capacitor stores due to its charge separation. The electric potential energy, denoted as \(U\), is the energy held in a charged capacitor and is directly proportional to the square of the charge \(Q\) held on the capacitor and inversely proportional to the capacitance \(C\). Essentially, it's the work done to move charge onto the plates against the electric field.

To put it in formulaic terms, the electric potential energy is calculated using the expression \(U = \frac{1}{2} C V^2\), where \(V\) represents the potential difference across the capacitor's plates, which, in the case of a spherical capacitor like Earth, is derived from the charge and the capacitance as \(V = \frac{Q}{C}\). Substituting this into the energy formula gives us \(U = \frac{1}{2} \frac{Q^2}{C}\).

Understanding how the Earth stores this energy provides us insight into natural phenomena such as lightning and tells us about the potential impacts on technology and human-made structures due to electrical discharges from the Earth's natural capacitor.
Capacitance Formula
The capacitance formula is fundamental to our understanding of how capacitors behave and function. Capacitance, represented by \(C\), measures a capacitor's ability to hold charge per unit potential difference, with the unit of measure being the farad (F). The general formula for capacitance is \(C = \frac{Q}{V}\), where \(Q\) is the charge stored on the capacitor, and \(V\) is the potential difference between the two conductors.

For different capacitor configurations, such as spherical, cylindrical, or parallel-plate capacitors, the capacitance formulas vary accordingly. As we've seen in the case of Earth as a spherical capacitor, the formula becomes \(C = 4\pi \epsilon_0 R\). Here, \(4\pi R\) reflects the surface area of the sphere which is relevant to the distribution of charge over a spherical surface.

Studying the capacitance formula enables students to predict how much charge a capacitor can store, which is vital in designing circuits and understanding the behavior of various electrical devices. A key point to remember is that the capacitance is independent of the actual amount of charge on the capacitor; it's a property of the geometry and materials making up the capacitor.

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Most popular questions from this chapter

A 4.00 -pF parallel plate capacitor has a potential difference of \(10.0 \mathrm{~V}\) across it. The plates are \(3.00 \mathrm{~mm}\) apart, and the space between them contains air. a) What is the charge on the capacitor? b) How much energy is stored in the capacitor? c) What is the area of the plates? d) What would the capacitance of this capacitor be if the space between the plates were filled with polystyrene?

A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is \(32.0 \mathrm{pF}\) when the separation between plates is \(d=0.500 \mathrm{~cm} .\) a) A battery with potential difference \(V=9.0 \mathrm{~V}\) is connected to the plates. What is the charge distribution, \(\sigma\), on the left plate? What are the capacitance, \(C^{\prime},\) and charge distribution, \(\sigma^{\prime},\) when \(d\) is changed to \(0.250 \mathrm{~cm} ?\) b) With \(d=0.500 \mathrm{~cm}\), the battery is disconnected from the plates. The plates are then moved so that \(d=0.250 \mathrm{~cm}\) What is the potential difference \(V^{\prime},\) between the plates?

Consider a cylindrical capacitor, with outer radius \(R\) and cylinder separation \(d\). Determine what the capacitance approaches in the limit where \(d \ll R\). (Hint: Express the capacitance in terms of the ratio \(d / R\) and then examine what happens as the ratio \(d / R\) becomes very small compared to \(1 .)\) Explain why the limit on the capacitance makes sense.

Determine all the values of equivalent capacitance you can create using any combination of three identical capacitors with capacitance \(C\).

You have an electric device containing a \(10.0-\mu \mathrm{F}\) capacitor, but an application requires an \(18.0-\mu \mathrm{F}\) capacitor. What modification can you make to your device to increase its capacitance to \(18.0-\mu \mathrm{F} ?\)

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