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In the air over a particular region at an altitude of \(500 \mathrm{m}\) above the ground the electric field is \(120 \mathrm{N} / \mathrm{C}\) directed downward. At 600 m above the ground the electric field is \(100 \mathrm{N} / \mathrm{C}\) downward. What is the average volume charge density in the layer of air between these two elevations? Is it positive or negative?

Short Answer

Expert verified
The average volume charge density in the layer of air between 500 m and 600 m above the ground is -3.5 脳 10^{-10} C/m鲁 and it is negative.

Step by step solution

01

Identify given variables

The electric field at an altitude of 500 m is 120 N/C, and at 600 m it is 100 N/C. The altitude difference is \(螖h = 600 m - 500 m = 100 m\).
02

Use Gauss's Law to derive a formula for volume charge density

According to Gauss's Law, the electric field \(E\) is given by \(E = \frac{\sigma}{2蔚_0}\) for a charged plane, where \(蟽\) is the charge density and \(蔚_0\) is the permittivity of free space. Because the charge is distributed through a volume, we use charge density \(\rho = \frac{\sigma}{螖h}\ The difference in electric fields is therefore given by \(螖E = E_2 - E_1 = \frac{\rho \Delta h}{2蔚_0}\). Solving for \(\rho\) gives \(\rho = \frac{2 蔚_0 \Delta E}{\Delta h}\).
03

Substitute the given values

Substitute given values into the formula: \( 蟻 = \frac{2 蔚_0 (E_1 - E_2)}{螖h}\). The permittivity of free space \(蔚_0\) is \(8.85 脳 10^{-12} C^2/N路m^2\), \(E_1\) is 120 N/C, \(E_2\) is 100 N/C and \(螖h\) is 100 m.
04

Calculate the average volume charge density

By plugging numbers into the formula, the average volume charge density is \( 蟻 = \frac{2 脳 8.85 脳 10^{-12} C^2/N路m^2 脳 (120 N/C - 100 N/C)}{100 m} = 3.5 脳 10^{-10} C/m鲁\).
05

Determine the charge's sign

The electric field points downward, which indicates that the direction of the force that a positive test charge would experience is toward the ground. This indicates that the air is negatively charged, so the average volume charge density is negative, 蟻 = -3.5 脳 10^{-10} C/m鲁.

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Key Concepts

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

Gauss's Law
Understanding Gauss's Law is crucial when studying electrostatics. It offers a practical way to calculate electric fields when dealing with symmetrical charge distributions. The law states that the net electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. Mathematically, it's expressed as \( \Phi = \frac{Q}{\epsilon_0} \) where \( \Phi \) is the electric flux, \( Q \) is the total charge enclosed, and \( \epsilon_0 \) is the permittivity of free space.

In simple terms, Gauss's Law connects the electric field surrounding a volume to the charge within that volume. In the context of an exercise, when you need to determine the volume charge density, you'd use Gauss's Law as a starting point. By considering a 'Gaussian surface' around the region of interest, you can analyze the change in the electric field and relate it to the charge density, leading to understanding how charge is distributed in a volume.
Electric Field
The electric field is a vector field around charged particles or objects that exerts forces on other charges within the field. Its strength is measured in newtons per coulomb (N/C), and it essentially reveals how a charge would behave if placed in the vicinity of another charge. The direction of the electric field gives the direction of the force that a positive test charge would experience.

For the example given, electric field measurements at different altitudes were used to infer the volume charge density of the air. The change in the value of the electric field at different heights implied that there was a change in the charge density. The downward direction of the field indicates that a positive test charge would be forced downward, which suggests the presence of negative charges in the atmosphere above.
Permittivity of Free Space
The permittivity of free space, denoted as \( \epsilon_0 \), is a fundamental constant that represents how electric fields interact with the vacuum of free space. Its value is approximately \( 8.85 \times 10^{-12} \, C^2/N\cdot m^2 \).

This constant appears in many equations in electromagnetism, such as in Coulomb's law and Gauss's Law, and it affects the strength of the electric field generated by a given charge. In calculations, \( \epsilon_0 \) can be seen as a scaling factor that determines how much influence a charge has on its surroundings; a greater \( \epsilon_0 \) would mean a comparatively weaker field for the same charge.
Volume Charge Density
Volume charge density is a measure of how much electric charge is distributed throughout a given volume. It is defined as the charge per unit volume and is represented by the symbol \( \rho \). In formulas, volume charge density is used to relate the amount of charge in a space to the resulting electric field and potential.

In the exercise you're reflecting on, the average volume charge density was determined by contrasting the electric field strength at two different altitudes, revealing how charge is spread out in a particular layer of the atmosphere. This measure helps in quantifying the amount of charge in a region, and it's essential to note that its sign is indicative of whether the charge is positive or negative. In the exercise, since the electric field is directed downward, the average volume charge density is negative, signifying an excess of electrons over positive ions in that atmospheric layer.

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Most popular questions from this chapter

A uniformly charged spherical shell with surface charge density \(\sigma\) contains a circular hole in its surface. The radius of the hole is small compared with the radius of the sphere. What is the electric field at the center of the hole? (Suggestion: This problem, like Problem \(64,\) can be solved by using the idea of superposition.)

A vertical electric field of magnitude \(2.00 \times 10^{4} \mathrm{N} / \mathrm{C}\) exists above the Earth's surface on a day when a thunderstorm is brewing. A car with a rectangular size of \(6.00 \mathrm{m}\) by \(3.00 \mathrm{m}\) is traveling along a roadway sloping downward at \(10.0^{\circ}\) Determine the electric flux through the bottom of the car.

A spherically symmetric charge distribution has a charge density given by \(\rho=a / r,\) where \(a\) is constant. Find the electric field as a function of \(r .\) (Suggestion: The charge within a sphere of radius \(R\) is equal to the integral of \(\rho d V\) where \(r\) extends from 0 to \(R\). To evaluate the integral, note that the volume element \(d V\) for a spherical shell of radius \(r\) and thickness \(d r\) is equal to \(4 \pi r^{2} d r\) .)

A point charge of \(12.0 \mu \mathrm{C}\) is placed at the center of a spherical shell of radius \(22.0 \mathrm{cm} .\) What is the total electric flux through (a) the surface of the shell and (b) any hemispherical surface of the shell? (c) Do the results depend on the radius? Explain.

An uncharged nonconducting hollow sphere of radius \(10.0 \mathrm{cm}\) surrounds a \(10.0-\mu \mathrm{C}\) charge located at the origin of a cartesian coordinate system. A drill with a radius of \(1.00 \mathrm{mm}\) is aligned along the \(z\) axis, and a hole is drilled in the sphere. Calculate the electric flux through the hole.

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