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Chapter 21: Problem 40

The earth is negatively charged, carrying \(500,000 \mathrm{C}\) of electric charge. This results in a \(300 \mathrm{kV}\) potential difference between the earth and the positively charged ionosphere. What is the capacitance of the earth-ionosphere system? If we assume that the bottom of the ionosphere is \(60 \mathrm{km}\) above the surface, what is the average electric field between the earth and the ionosphere?

Short Answer

Expert verified
The capacitance of the earth-ionosphere system is approximately \(1.67 \mu F\), and the average electric field between them is approximately \(5.0 \mathrm{kV/m}\).

Step by step solution

01

Calculate the Capacitance

First, gather the given variables. The electric charge (\(Q\)) is \(500,000 \mathrm{C}\) and the potential difference (\(V\)) is \(300 \mathrm{kV}\), which is equivalent to \(300,000 \mathrm{V}\). Then, substitute these values into the capacitance formula \(C = Q/V\). After performing the calculation, the capacitance (\(C\)) can be determined.
02

Calculate the Electric Field

In this step, gather the given variables. The potential difference (\(V\)) is still \(300,000 \mathrm{V}\), and the distance (\(d\)) from the surface of the Earth to the ionosphere is \(60 \mathrm{km}\), which is equivalent to \(60,000 \mathrm{m}\). Substitute these values into the electric field formula \(E = V/d\). By performing the calculation, the electric field (\(E\)) can be found.

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Key Concepts

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

Understanding Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. There are two types of electric charges: positive and negative. Like charges repel each other, while opposite charges attract. The standard unit of electric charge is the coulomb (C).

In the exercise, the Earth carries a negative charge of 500,000 C. This large quantity of charge influences the electric field and potential difference in the Earth-ionosphere system. It is the imbalance of electric charge between the Earth and the ionosphere that allows us to calculate the system's capacitance and the electric field strength between them.
Potential Difference Explained
The potential difference, commonly referred to as voltage, is the measure of the work needed to move a charge from one point to another in an electric field. It is the energy per unit charge and is measured in volts (V). A potential difference of 1 V means that 1 joule of work is required to move a charge of 1 C between two points.

The textbook example mentions a 300 kV (or 300,000 V) potential difference between the Earth and the ionosphere. This value signifies the energy required to move charge between these two points. This potential difference is vital for calculating the capacitance of the Earth-ionosphere system. By dividing the electric charge by the potential difference, as done in the step-by-step solution, we can find the capacitance, which represents the system's ability to store electrical energy.
Electric Field Characteristics
An electric field is a region around a charged object where forces are exerted on other charged objects. The strength of the electric field is directly related to the charge that creates it and inversely related to the square of the distance from the charge. Electric field strength is measured in volts per meter (V/m).

When calculating the average electric field between the Earth and the ionosphere, as shown in the solution, the potential difference (V) is divided by the distance (d) over which the field is exerted. The resulting electric field describes the effect of the potential difference over that specified distance. Not only does this concept help us understand the forces experienced by charges in the system, but it also gives us insight into how energy is distributed within the electric field.

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Most popular questions from this chapter

Investigators are exploring ways to treat milk for longer shelf life by using pulsed electric fields to destroy bacterial contamination. One system uses 8.0 -cm-diameter circular plates separated by \(0.95 \mathrm{cm} .\) The space between the plates is filled with milk, which has a dielectric constant the same as that of water. The plates are briefly charged to 30,000 V. What is the capacitance of the system, and how much charge is on each plate when they are fully charged?

Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio \(\left(U_{\mathrm{C}}\right)_{1} /\left(U_{\mathrm{C}}\right)_{2} ?\)

A Na' ion moves from inside a cell, where the electric potential is \(-70 \mathrm{mV},\) to outside the cell, where the potential is \(0 \mathrm{V}\). What is the change in the ion's electric potential energy as it moves from inside to outside the cell? Does its energy increase or decrease?

To provide the pulse of energy needed for an intense bass, some car stereo systems add capacitors. One system uses a 2.0 F capacitor charged to 24 V, double the normal 12 V provided by the car's battery. How much energy does the capacitor store at \(12 \mathrm{V}\) ? At \(24 \mathrm{V} ?\)

In the early 1900 s, Robert Millikan used small charged droplets of oil, suspended in an electric field, to make the first quantitative measurements of the electron's charge. A 0.70- \(\mu \mathrm{m}\) -diameter droplet of oil, having a charge of \(+e\), is suspended in midair between two horizontal plates of a parallel-plate capacitor. The upward electric force on the droplet is exactly balanced by the downward force of gravity. The oil has a density of \(860 \mathrm{kg} / \mathrm{m}^{3},\) and the capacitor plates are \(5.0 \mathrm{mm}\) apart. What must the potential difference between the plates be to hold the droplet in equilibrium?
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