/*! 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

An insulating hollow sphere has inner radius \(a\) and outer radius \(b\). Within the insulating material the volume charge density is given by \(\rho\) (\(r\)) \(= \alpha/r\), where \(\alpha\) is a positive constant. (a) In terms of \(\alpha\) and \(a\), what is the magnitude of the electric field at a distance \(r\) from the center of the shell, where \(a < r < b\)? (b) A point charge \(q\) is placed at the center of the hollow space, at \(r =\) 0. In terms of \(\alpha\) and \(a\), what value must \(q\) have (sign and magnitude) in order for the electric field to be constant in the region \(a < r < b\), and what then is the value of the constant field in this region?

Which statement is true about \(\overrightarrow{E}\) inside a negatively charged sphere as described here? (a) It points from the center of the sphere to the surface and is largest at the center. (b) It points from the surface to the center of the sphere and is largest at the surface. (c) It is zero. (d) It is constant but not zero.

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.

In one experiment the electric field is measured for points at distances \(r\) from a uniform line of charge that has charge per unit length \(\lambda\) and length \(l\), where \(l \gg r\). In a second experiment the electric field is measured for points at distances \(r\) from the center of a uniformly charged insulating sphere that has volume charge density \(\rho\) and radius \(R =\) 8.00 mm, where \(r > R\). The results of the two measurements are listed in the table, but you aren't told which set of data applies to which experiment: For each set of data, draw two graphs: one for \(Er^2\) versus r and one for \(Er\) versus \(r\). (a) Use these graphs to determine which data set, A or B, is for the uniform line of charge and which set is for the uniformly charged sphere. Explain your reasoning. (b) Use the graphs in part (a) to calculate \(\lambda\) for the uniform line of charge and \(\rho\) for the uniformly charged sphere.

It was shown in Example 21.10 (Section 21.5) that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude \(E = \lambda/2\pi\varepsilon_0r\). Consider an imaginary cylinder with radius \(r =\) 0.250 m and length \(l =\) 0.400 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is \(\lambda =\) 3.00 \(\mu\)C/m. (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to \(r =\) 0.500 m? (c) What is the flux through the cylinder if its length is increased to \(l =\) 0.800 m?

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