A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area σ=5.00×10−6\sigma=5.00\$$times10^{-6}$$σ=5.00×10−6 C/m2. A small sphere of mass m=8.00×10−6m=8.00\$$times10^{-6}m=8.00×10$$−6 kg and charge qq is placed 3.00 3.003.00 cm above the sheet of charge and then released from rest. What is qq if the sphere is released 1.501.50 cm above the sheet?

Ch 22: Gauss' Law

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 22: Gauss' Law

문제 26b

22장, 문제 26b

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?

검증된 단계별 안내

Step 1: Understand the problem setup. We have a large sheet with a uniform charge density $ \sigma = 5.00 \times 10^{-6} \text{ C/m}^2 $. A small sphere with mass $ m = 8.00 \times 10^{-6} \text{ kg} $ and charge $ q $ is placed above the sheet. We need to find the charge $ q $ when the sphere is released from different heights above the sheet.

Step 2: Use Gauss's Law to find the electric field $ E $ due to the sheet. For an infinite sheet of charge, the electric field is given by $ E = \frac{\sigma}{2\varepsilon_0} $, where $ \varepsilon_0 $ is the permittivity of free space $ \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 $.

Step 3: Calculate the force $ F $ acting on the sphere due to the electric field. The force is given by $ F = qE $. Since the sphere is released from rest, this force will cause it to accelerate towards the sheet.

Step 4: Apply Newton's second law to relate the force to the acceleration $ a $ of the sphere. According to Newton's second law, $ F = ma $, where $ m $ is the mass of the sphere. Therefore, $ qE = ma $.

Step 5: Solve for the charge $ q $. Rearrange the equation $ qE = ma $ to find $ q = \frac{ma}{E} $. Substitute the values for $ m $, $ a $, and $ E $ to find $ q $. Note that the acceleration $ a $ can be determined from the change in height and the initial conditions of the sphere's motion.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Electric Field due to a Charged Sheet

The electric field generated by an infinite sheet of charge is uniform and perpendicular to the surface. It is given by E = σ / (2ε₀), where σ is the charge density and ε₀ is the permittivity of free space. This field does not depend on the distance from the sheet, making it crucial for understanding the forces acting on nearby charged objects.

Force on a Charged Particle in an Electric Field

A charged particle in an electric field experiences a force given by F = qE, where q is the charge of the particle and E is the electric field strength. This force influences the motion of the particle, and understanding it is essential for predicting how the sphere will move when released above the charged sheet.

Gravitational Force

The gravitational force acting on an object is given by F = mg, where m is the mass of the object and g is the acceleration due to gravity. In this scenario, the gravitational force opposes the electric force, and analyzing the balance between these forces is key to determining the charge q required for equilibrium or specific motion.