A hemispherical surface with radius rr in a region of uniform electric field E→\overrightarrow{E} has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

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 5

Chapter 22, Problem 5

A hemispherical surface with radius $r$ in a region of uniform electric field $\vec{E}$ has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

Verified step by step guidance

Understand that electric flux (Φ) through a surface is given by the formula Φ = ∫E→ · dA→, where E→ is the electric field and dA→ is the differential area vector.

Recognize that the hemispherical surface is aligned such that its axis is parallel to the electric field E→. This means the electric field is perpendicular to the flat circular base of the hemisphere and parallel to the curved surface.

Calculate the flux through the flat circular base of the hemisphere. Since the electric field is perpendicular to this surface, the flux through the base is Φ_base = E * A_base, where A_base = πr² is the area of the circular base.

Consider the flux through the curved surface of the hemisphere. The electric field lines are parallel to this surface, meaning the angle between the electric field and the area vector is 90 degrees, resulting in zero flux through the curved surface.

Combine the flux contributions: The total flux through the hemispherical surface is simply the flux through the flat base, as the flux through the curved surface is zero. Therefore, Φ_total = Φ_base = E * πr².

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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 quantifies the amount of electric field passing through a surface. It is calculated as the dot product of the electric field vector and the area vector, integrating over the surface. For a uniform field and flat surface, it simplifies to the product of the field strength, area, and the cosine of the angle between them.

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 is equal to the enclosed charge divided by the permittivity of free space. This principle is useful for calculating flux in symmetrical situations, such as spherical or cylindrical geometries.

Surface Area of a Hemisphere

The surface area of a hemisphere is crucial for calculating flux through it. A hemisphere's curved surface area is half that of a full sphere, given by 2πr², where r is the radius. This area is used in conjunction with the electric field to determine the total flux through the hemisphere.

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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. 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.

The nuclei of large atoms, such as uranium, with $92$ protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately $7.4 \times 10^{- 15}$ m. What is the electric field this nucleus produces just outside its surface?

You measure an electric field of $1.25 \times 10^{6}$ 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

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter $12.0$ cm, giving it a charge of $−49.0$ $μ$ C. Find the electric field just outside the paint layer;

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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?

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter $12.0$ cm, giving it a charge of $negative 49.0$ $μ$ C. Find the electric field just inside the paint layer.

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A hemispherical surface with radius rr in a region of uniform electric field E→\(\overrightarrow{E}\) has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

Verified video answer for a similar problem:

Key Concepts

Electric Flux

Gauss's Law

Surface Area of a Hemisphere

A hemispherical surface with radius $r$ in a region of uniform electric field $\vec{E}$ has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.