/*! 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 26 A very large, horizontal, noncon... [FREE SOLUTION] | 91Ó°ÊÓ

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A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area \(\sigma =\) 5.00 \(\times\) 10\(^{-6}\) C/m\(^2\). (a) A small sphere of mass \(m =\) 8.00 \(\times\) 10\(^{-6}\) kg and charge \(q\) is placed 3.00 cm above the sheet of charge and then released from rest. (a) If the sphere is to remain motionless when it is released, what must be the value of \(q\)? (b) What is \(q\) if the sphere is released 1.50 cm above the sheet?

Short Answer

Expert verified
(a) \( q \approx 2.78 \times 10^{-17} \) C; (b) Same \( q \) value because the field is uniform.

Step by step solution

01

Identify Forces on Sphere

The sphere experiences two main forces: the electric force due to the charged sheet and the gravitational force due to its mass. The electric force needs to balance the gravitational force for the sphere to remain motionless.
02

Express Electric Force

The electric field due to a large charged sheet with surface charge density \( \sigma \) is \( E = \frac{\sigma}{2\varepsilon_0} \). The force on the sphere is then \( F_{\text{electric}} = qE = q \frac{\sigma}{2\varepsilon_0} \).
03

Express Gravitational Force

The gravitational force acting on the sphere is simply its weight: \( F_{\text{gravity}} = mg \), where \( m = 8.00 \times 10^{-6} \) kg and \( g = 9.81 \text{ m/s}^2 \).
04

Equate Forces for Equilibrium

For the sphere to remain motionless, the electric force must equal the gravitational force. Thus, \( q \frac{\sigma}{2\varepsilon_0} = mg \). Rearrange to find \( q = \frac{2mg\varepsilon_0}{\sigma} \).
05

Calculate Value of q for 3 cm

Substitute the values into the equation: \( q = \frac{2 \times 8.00 \times 10^{-6} \times 9.81 \times 8.85 \times 10^{-12}}{5.00 \times 10^{-6}} \). Calculate to find \( q \approx 2.78 \times 10^{-17} \) C.
06

Analyze Impact of Height

The height (3.00 cm or 1.50 cm) does not affect the values calculated since the electric field from an infinite sheet is constant regardless of distance, hence the same calculation applies for any height.

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Key Concepts

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

Nonconducting Sheet
A nonconducting sheet is a surface that doesn't allow the flow of electric charge across it. This kind of sheet usually holds a static charge. In this exercise, we're dealing with a large, horizontal nonconducting sheet. Such sheets are particularly interesting in electromagnetics because they create a uniform electric field around them.
When a charge is spread evenly on the surface of a nonconducting sheet, it creates an electric field that is perpendicular to the sheet's surface. The field's intensity depends on the sheet's surface charge density, and this field extends into the space around the sheet.
Imagine our exercise scenario, with a sphere placed above the sheet. The electric field generated by the sheet influences the sphere, regardless of its vertical position. So, a key takeaway is that for very large sheets, the field strength does not diminish with distance from the sheet.
Surface Charge Density
Surface charge density, represented by the Greek letter sigma (\( \sigma \)), is a measure of how much electric charge is distributed over a surface. It's defined mathematically as charge per unit area, with units of coulombs per square meter (C/m²).
In this problem, the value provided is \( \sigma = 5.00 \times 10^{-6} \text{ C/m}^2 \). This number tells us how densely packed the charge is on the nonconducting sheet. The higher the surface charge density, the stronger the electric field produced by the sheet.
Surface charge density is crucial for calculating the electric field and, subsequently, the electric force on other charged objects nearby. For our exercise, it helps establish the strength of the electric field that the sphere interacts with when positioned close to the sheet.
Equilibrium of Forces
In physics, equilibrium refers to a state where all forces acting on an object are balanced, resulting in no net force and no acceleration. So, the object remains at rest or in constant motion.
For the sphere in our exercise, equilibrium is achieved when the electric force due to the nonconducting sheet equals and opposes the gravitational force pulling it downward. This balance keeps the sphere motionless.
To find this point of equilibrium, we set the equation for the electric force equal to the gravitational force: \( q \frac{\sigma}{2\varepsilon_0} = mg \). Solving this helps us determine the specific charge \( q \) the sphere must have to remain stationary. This kind of equilibrium of forces is a classic illustration of Newton's first law of motion.
Charged Sphere
A charged sphere is a small object that carries an electric charge. In this exercise, the sphere has mass \( m = 8.00 \times 10^{-6} \text{ kg} \) and charge \( q \). When placed in an electric field, the sphere experiences a force proportional to its charge.
The charged sphere's position above the nonconducting sheet is important as it allows us to analyze the effects of the electric field generated by the surface charge density. We seek to find the exact amount of charge \( q \) required for the sphere to remain motionless in such a field.
Interestingly, for an infinite nonconducting sheet, the electric field it produces doesn't change with distance, so whether the sphere is 3 cm or 1.5 cm away does not affect the field strength or the force experienced by the sphere. This aspect simplifies many calculations when considering large charged surfaces.

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Most popular questions from this chapter

A solid conducting sphere carrying charge \(q\) has radius \(a\). It is inside a concentric hollow conducting sphere with inner radius \(b\) and outer radius \(c\). The hollow sphere has no net charge. (a) Derive expressions for the electricfield magnitude in terms of the distance \(r\) from the center for the regions \(r < a, a < r < b, b < r < c\), and \(r > c\). (b) Graph the magnitude of the electric field as a function of \(r\) from \(r =\) 0 to \(r =\) 2c. (c) What is the charge on the inner surface of the hollow sphere? (d) On the outer surface? (e) Represent the charge of the small sphere by four plus signs. Sketch the field lines of the system within a spherical volume of radius 2\(c\).

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes \(x = d\) and \(x = -d\). The \(y\)- and \(z\)-dimensions of the slab are very large compared to \(d\); treat them as essentially infinite. The slab has a uniform positive charge density \(\rho\). (a) Explain why the electric field due to the slab is zero at the center of the slab (\(x =\) 0). (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space.

(a) At a distance of 0.200 cm from the center of a charged conducting sphere with radius 0.100 cm, the electric field is 480 N/C. What is the electric field 0.600 cm from the center of the sphere? (b) At a distance of 0.200 cm from the axis of a very long charged conducting cylinder with radius 0.100 cm, the electric field is 480 N/C. What is the electric field 0.600 cm from the axis of the cylinder? (c) At a distance of 0.200 cm from a large uniform sheet of charge, the electric field is 480 N/C. What is the electric field 1.20 cm from the sheet?

Early in the 20th century, a leading model of the structure of the atom was that of English physicist J. J. Thomson (the discoverer of the electron). In Thomson's model, an atom consisted of a sphere of positively charged material in which were embedded negatively charged electrons, like chocolate chips in a ball of cookie dough. Consider such an atom consisting of one electron with mass \(m\) and charge \(-e\), which may be regarded as a point charge, and a uniformly charged sphere of charge \(+e\) and radius \(R\). (a) Explain why the electron's equilibrium position is at the center of the nucleus. (b) In Thomson's model, it was assumed that the positive material provided little or no resistance to the electron's motion. If the electron is displaced from equilibrium by a distance less than \(R\), show that the resulting motion of the electron will be simple harmonic, and calculate the frequency of oscillation. (\(Hint:\) Review the definition of SHM in Section 14.2. If it can be shown that the net force on the electron is of this form, then it follows that the motion is simple harmonic. Conversely, if the net force on the electron does not follow this form, the motion is not simple harmonic.) (c) By Thomson's time, it was known that excited atoms emit light waves of only certain frequencies. In his model, the frequency of emitted light is the same as the oscillation frequency of the electron(s) in the atom. What radius would a Thomson-model atom need for it to produce red light of frequency 4.57 \(\times\) 10\(^{14}\) Hz? Compare your answer to the radii of real atoms, which are of the order of 10\(^{-10}\) m (see Appendix F). (d) If the electron were displaced from equilibrium by a distance greater than \(R\), would the electron oscillate? Would its motion be simple harmonic? Explain your reasoning. (\(Historical\) \(note:\) In 1910, the atomic nucleus was discovered, proving the Thomson model to be incorrect. An atom's positive charge is not spread over its volume, as Thomson supposed, but is concentrated in the tiny nucleus of radius 10\(^{-14}\) to 10\(^{-15}\) m.)

A solid metal sphere with radius 0.450 m carries a net charge of 0.250 nC. Find the magnitude of the electric field (a) at a point 0.100 m outside the surface of the sphere and (b) at a point inside the sphere, 0.100 m below the surface.

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