A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area σ=5.00×10−6\sigma=5.00\$$times10^{-6} C/m2. A $$small sphere of mass m=8.00×10−6m=8.00\$$times10^{-6} kg $$and charge qq is placed 3.00 3.00 cm above the sheet of charge and then released from rest. If the sphere is to remain motionless when it is released, what must be the value of qq?

Ch 22: Gauss' Law

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 22: Gauss' Law

Problem 26a

Chapter 22, Problem 26a

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area $σ = 5.00 \times 10^{- 6}$ C/m². 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. If the sphere is to remain motionless when it is released, what must be the value of $q$ ?

Verified step by step guidance

Understand the problem: We have a charged sheet and a charged sphere. The sphere is released from rest and should remain motionless. This implies that the forces acting on the sphere must be balanced.

Identify the forces: The sphere experiences two forces: the gravitational force pulling it downward and the electric force due to the charged sheet pushing it upward.

Calculate the gravitational force: The gravitational force $ F_{gravity} $ acting on the sphere is given by $ F_{gravity} = m \cdot g $, where $ m $ is the mass of the sphere and $ g $ is the acceleration due to gravity (approximately $ 9.81 \text{ m/s}^2 $).

Calculate the electric force: The electric field $ E $ due to a large sheet of charge is given by $ E = \frac{\sigma}{2\varepsilon_0} $, where $ \sigma $ is the charge per unit area and $ \varepsilon_0 $ is the permittivity of free space (approximately $ 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 $). The electric force $ F_{electric} $ on the sphere is $ F_{electric} = q \cdot E $.

Set the forces equal for equilibrium: For the sphere to remain motionless, the electric force must equal the gravitational force. Therefore, $ q \cdot E = m \cdot g $. Substitute the expression for $ E $ and solve for $ q $: $ q = \frac{m \cdot g}{\frac{\sigma}{2\varepsilon_0}} $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field of a Charged Sheet

The electric field generated by a large, uniformly charged nonconducting sheet is constant and perpendicular to the sheet. It is given by E = σ / (2ε₀), where σ is the charge density and ε₀ is the permittivity of free space. This field influences the motion of charged objects placed near the sheet.

Force Balance

For the sphere to remain motionless, the net force acting on it must be zero. This requires balancing the gravitational force (mg) acting downward with the electric force (qE) acting upward. Solving for q involves equating these forces, ensuring the sphere remains stationary.

Gravitational Force

The gravitational force acting on an object is given by F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.81 m/s² on Earth). This force acts downward and must be counteracted by the electric force for the sphere to remain motionless.

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Related Practice

Textbook Question

Charge $q$ is distributed uniformly throughout the volume of an insulating sphere of radius $R equals 4.00$ cm. At a distance of $r equals 8.00$ cm from the center of the sphere, the electric field due to the charge distribution has magnitude $E equals 940$ N/C. What is the volume charge density for the sphere?

A conductor with an inner cavity, like that shown in Fig. $22.23$ c, carries a total charge of $positive 5.00$ nC. The charge within the cavity, insulated from the conductor, is $negative 6.00$ nC. How much charge is on (a) the inner surface of the conductor and (b) the outer surface of the conductor?

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Textbook Question

An infinitely long cylindrical conductor has radius $r$ and uniform surface charge density $σ$ . In terms of $σ$ , what is the magnitude of the electric field produced by the charged cylinder at a distance $r > R$ from its axis? Then, express the result in terms of $λ$ and show that the electric field outside the cylinder is the same as if all the charge were on the axis.

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area $$\sigma$=5.00$\times$10^{-6}$ C/m². 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. What is $q$ if the sphere is released $1.50$ cm above the sheet?

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Textbook Question

Charge $q$ is distributed uniformly throughout the volume of an insulating sphere of radius $R = 4.00$ cm. At a distance of $r = 8.00$ cm from the center of the sphere, the electric field due to the charge distribution has magnitude $E = 940$ N/C. What is the electric field at a distance of $2.00$ cm from the sphere's center?

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Textbook Question

An infinitely long cylindrical conductor has radius $r$ and uniform surface charge density $σ$ . In terms of $σ$ and $R$ , what is the charge per unit length $λ$ for the cylinder?

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