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Ch 22: Gauss' Law
Chapter 22, Problem 22

A hemispherical surface with radius r in a region of uniform electric field E→ has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.Diagram of a hemispherical surface in a uniform electric field, illustrating electric flux.

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Identify the given parameters: the radius of the hemispherical surface (r) and the uniform electric field (E).
Recall the formula for electric flux, which is given by Φ = E * A * cos(θ), where A is the area through which the field lines pass and θ is the angle between the field lines and the normal to the surface.
Note that the electric field is parallel to the axis of the hemisphere, so θ = 0 degrees, and cos(0) = 1.
Calculate the area of the flat circular surface of the hemisphere, which is A = π * r^2.
Substitute the values into the flux formula: Φ = E * π * r^2.

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Key Concepts

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

Electric Flux

Electric flux is a measure of the quantity of electric field lines passing through a given surface. It is mathematically defined as the dot product of the electric field vector and the area vector of the surface, integrated over the surface. The formula for electric flux (Φ) is given by Φ = ∫ E · dA, where E is the electric field and dA is the differential area element. Understanding electric flux is crucial for solving problems related to electric fields and Gauss's law.
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Electric Flux

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε₀). This law is fundamental in electrostatics and simplifies the calculation of electric fields for symmetric charge distributions. It provides a powerful tool for analyzing electric fields in various geometries.
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Uniform Electric Field

A uniform electric field is one in which the electric field strength is constant in both magnitude and direction throughout a region. This means that the field lines are parallel and evenly spaced. In the context of the problem, the hemispherical surface is placed in such a field, which simplifies the calculation of electric flux since the angle between the electric field and the surface normal can be easily determined. Understanding the characteristics of uniform electric fields is essential for analyzing their effects on charged objects.
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Intro to Electric Fields
Related Practice
Textbook Question
A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37×10−6 C/m2. A charge of −0.500 μC is now introduced at the center of the cavity inside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?
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Textbook Question
A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not?
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Textbook Question
A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (c) For what angle φ between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.
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Textbook Question
You measure an electric field of 1.25×106 N/C at a distance of 0.150 m from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.150 m? (b) What is the magnitude of this charge?
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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 are (a) the volume charge density for the sphere?
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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 are (b) the electric field at a distance of 2.00 cm from the sphere's center?
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