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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.

Short Answer

Expert verified
The flux through the hemispherical surface is \( E \pi r^2 \).

Step by step solution

01

Understanding Electric Flux

Electric flux through a surface is a measure of how much the electric field penetrates through that surface. It is calculated using the formula: \( \Phi = \int \vec{E} \cdot d\vec{A} \), where \( \vec{E} \) is the electric field vector and \( d\vec{A} \) is the differential area vector.
02

Identify the Geometry

The problem involves a hemispherical surface with a radius \( r \), aligned parallel to a uniform electric field \( \vec{E} \). The hemisphere implies that its curved surface and circular base must both be considered regarding flux calculation.
03

Flux Through Curved Surface

On the curved surface of the hemisphere, the electric field lines are perpendicular (normal) to the surface. Therefore, the flux through the curved surface is zero: \( \Phi_{curved} = \int \vec{E} \cdot d\vec{A} = 0 \).
04

Flux Through Circular Base

The flux through the flat, circular base, which is perpendicular to the electric field, is calculated by \( \Phi_{base} = E \cdot A \), where \( A = \pi r^2 \) is the area of the circular base. Thus, \( \Phi_{base} = E \pi r^2 \).
05

Total Flux Calculation

The total flux through the hemispherical surface is the sum of the flux through the curved surface and the circular base. Since \( \Phi_{curved} = 0 \), the total flux is simply \( \Phi_{total} = \Phi_{base} = E \pi r^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Hemispherical Surfaces
A hemispherical surface is essentially half of a sphere, much like a dome. When considering flux calculations, it becomes crucial to consider both its attributes: the curved surface and the flat, circular base. The curved surface represents only a portion of a complete sphere, and in mathematical terms, it comprises all the points that lie within a certain radius from a central point, but only for one hemisphere.

When you align the axis of this hemisphere with an electric field, you need to evaluate the effect of this orientation on how much electric field passes through these surfaces. To solve problems regarding hemispherical surfaces, it helps to remember:
  • There are two main parts—the curved top and the flat base.
  • The orientation with relation to vectors like the electric field affects which part allows more field lines to pass through.
  • The hemisphere's alignment can change the nature of the orthogonal (perpendicular) connections between the field and the surface.
In electrical flux problems, distinguishing between these parts can simplify the solutions and avoid errors in calculations.
Understanding Uniform Electric Fields
In a uniform electric field, the field lines are parallel and equally spaced. This implies the electric field strength is constant both in magnitude and direction at every point in the field. Such fields are described by a constant vector denoted as \( \vec{E} \).

Uniform electric fields are ideal for theoretical environments and make calculations easier because:
  • Every segment of a flat surface parallel to the field experiences the same electric field intensity.
  • The direction of the field is uniform, simplifying integration processes involved in flux calculations.
  • For hemispheres, aligning the field lines parallel to the symmetry axis aids in analytically determining the field interactions, simplifying the geometry consideration needed.
An illustrative example is charged parallel plates, where between them exists a uniform field. Understanding these fields is crucial since they present one of the more straightforward scenarios for calculating something like electric flux.
Flux Calculation for Hemispheres
Electric flux \( \Phi \) is a concept that denotes the number of electric field lines passing through a certain area. It combines the field strength and the surface geometry into a single quantity.

For a hemispherical surface in a uniform electric field, calculating the flux requires examining both the curved surface and the flat circular base:
  • Curved Surface: Here, the electric field vectors are parallel to the surface, implying no perpendicular component. Therefore, this contributes zero flux ( \( \Phi_{curved} = 0 \).
  • Circular Base: This part operates at 90 degrees to the field direction, allowing maximum field line penetration. The flux is determined by the product of the electric field magnitude \( E \) and the base area \( A = \pi r^2 \), giving us \( \Phi_{base} = E \pi r^2 \).

Ultimately, the total flux through the hemisphere is the sum of these individual fluxes: \( \Phi_{total} = \Phi_{curved} + \Phi_{base} = E \pi r^2 \). By simplifying the problem into manageable parts, hemispherical flux calculations become straightforward, allowing the aggregate flux to be computed effectively.

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Most popular questions from this chapter

A small conducting spherical shell with inner radius \(a\) and outer radius \(b\) is concentric with a larger conducting spherical shell with inner radius \(c\) and outer radius \(d\) (Fig. P22.47). The inner shell has total charge \(+2 q,\) and the outer shell has charge \(+4 q .\) (a) Calculate the electric field (magnitude and direction) in terms of \(q\) and the distance \(r\) from the common center of the two shells for (i) \(r < a ;\) (ii) \(a < r < b ;\) (iii) \(b < r < c ;\) (iv) \(c < r < d\) ; (v) \(r > d .\) Show your results in a graph of the radial component of \(\vec{\boldsymbol{E}}\) as a function of \(r .\) (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (ii) inner surface of the large shell; (iv) outer surface of the large shell?

A negative charge \(-Q\) is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not? (c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge \(-Q ?\) Is it reasonable to say that the grounded conductor has shielded the region from the effects of the charge \(-Q ?\) In principle, could the same thing be done for gravity? Why or why not?

Negative charge \(-Q\) is distributed uniformly over the surface of a thin spherical insulating shell with radius \(R .\) Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located (a) a distance \(r > R\) from the center of the shell (outside the shell) and (b) a distance \(r < R\) from the center of the shell (inside the shell).

The Coaxial Cable. A long coaxial cable consists of an inner cylindrical conductor with radius \(a\) and an outer coaxial cylinder with inner radius \(b\) and outer radius \(c .\) The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length \lambda. Calculate the electric field (a) at any point between the cylinders a distance \(r\) from the axis and (b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance \(r\) from the axis of the cable, from \(r=0\) to \(r=2 c\) . (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{m} .\) What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) \(0.500 \mathrm{m},(\mathrm{b}) 1.50 \mathrm{m},(\mathrm{c}) 2.50 \mathrm{m} ?\)

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