/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Field from a hemisphere ** (a)... [FREE SOLUTION] | 91Ó°ÊÓ

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Field from a hemisphere ** (a) What is the electric field at the center of a hollow hemispherical shell with radius \(R\) and uniform surface charge density \(\sigma\) ? (This is a special case of Problem \(1.12\), but you can solve the present exercise much more easily from scratch, without going through all the messy integrals of Problem 1.12.) (b) Use your result to show that the electric field at the center of a solid hemisphere with radius \(R\) and uniform volume charge density \(\rho\) equals \(\rho R / 4 \epsilon_{0}\)

Short Answer

Expert verified
For a hollow hemisphere, the electric field at its center is \(\frac{\sigma}{2 \epsilon_{0}}\). For a solid hemisphere, the electric field at its center is \(\frac{\rho R}{4 \epsilon_{0}}\).

Step by step solution

01

Use Gauss's Law for a hollow hemisphere

To find the electric field at the center of a hollow hemisphere with radius \(R\) and uniform surface charge density \(\sigma\), we must use Gauss's Law because the charge distribution is symmetrical. According to Gauss's Law, the electric field \(E\) is given by: \(\frac{Q_{\text{enc}}}{\epsilon_{0} A}\), where \(Q_{\text{enc}}\) is the enclosed electric charge, \(\epsilon_{0}\) is the electric constant, and \(A\) is the area of our Gaussian surface. The enclosed charge \(Q_{\text{enc}}\) is equal to the surface charge density \(\sigma\) times the surface area of the hemisphere, \(2\pi R^2\), so \(Q_{\text{enc}} = \sigma(2 \pi R^2)\). The electric field \(E\) then simplifies to \(\frac{\sigma(2\pi R^2)}{\epsilon_{0}(4\pi R^2)} = \frac{\sigma}{2 \epsilon_{0}}\).
02

Use Gauss's law for a solid hemisphere

Now we need to find the electric field at the center of a solid hemisphere with radius \(R\) and uniform volume charge density \(\rho\). Here, the Gaussian surface is still a sphere of radius \(R\), but now the enclosed charge \(Q_{\text{enc}}\) is equal to the volume charge density \(\rho\) times the volume of the hemisphere \(\frac{2}{3}\pi R^3\), therefore \(Q_{\text{enc}} = \rho(\frac{2}{3}\pi R^3)\). Using Gauss's Law, we can use these values to find the electric field \(E\) is \(\frac{\rho(2 \pi R^3 /3)}{\epsilon_{0}(4 \pi R^2)} = \frac{\rho R}{4 \epsilon_{0}}\), which is what we were supposed to show.

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Key Concepts

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

Gauss's Law
Gauss's Law plays a fundamental role in understanding electric fields. It states that the electric flux through any closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space. Mathematically, it's expressed as \( \phi_E = \frac{Q_{\text{enc}}}{\( \epsilon_{0} \)} \), where \( \phi_E \) is the electric flux, \( Q_{\text{enc}} \) is the enclosed charge, and \( \epsilon_{0} \) is the electric constant.

When dealing with symmetrical charge distributions like a hemisphere, Gauss's Law simplifies the complex integrations often required. Applying this law requires selecting an appropriate Gaussian surface that takes advantage of the symmetry, enabling the simplification of electric field calculations.
Surface Charge Density
Surface charge density, typically denoted as \( \sigma \), represents the amount of electric charge per unit area on the surface of a charged object. In the case of a hemisphere with uniform charge distribution, \( \sigma \) remains constant across the entire surface. This uniformity greatly aids in simplifying calculations; for instance, if you know \( \sigma \) and the surface area of the hemisphere \( (2\pi R^2) \), you can easily find the total charge \( Q_{\text{enc}} = \sigma(2 \pi R^2) \) on the hemisphere.
Volume Charge Density
Volume charge density, indicated as \( \rho \) is the charge per unit volume within a three-dimensional material. For a solid hemisphere with uniform charge distribution, \( \rho \) is constant throughout its volume. Determining the total enclosed charge \( Q_{\text{enc}} \) for such objects involves multiplying \( \rho \) by the object's volume. For a hemisphere, this enclosed charge is \( \rho(\frac{2}{3}\pi R^3) \) since the volume of a hemisphere is \( \frac{2}{3}\pi R^3 \).
Electric Constant
The electric constant, \( \epsilon_{0} \), also known as the permittivity of free space, is a fundamental constant in electromagnetism that describes the ability of a vacuum to permit electric field lines. This value is crucial in Gauss's Law, as it relates the amount of charge to the resulting electric field. It is often used in conjunction with surface or volume charge densities to calculate the electric field produced by various charge distributions, as seen in the exercise where the electric field is proportionally inverse to \( \epsilon_{0} \).
Uniform Charge Distribution
Uniform charge distribution implies that the charge is spread out evenly over a surface (surface charge density) or throughout a volume (volume charge density). This uniformity simplifies calculating electric fields, as it guarantees that the charge density remains constant, whether considering a point on the surface or any point within the volume. In our hemisphere example, both surface and volume charge densities are uniform, making the calculations for electric field straightforward using Gauss's Law.

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Most popular questions from this chapter

\(N\) charges on a circle \(N\) point charges, each with charge \(Q / N\), are evenly distributed around a circle of radius \(R\). What is the electric field at the location of one of the charges, due to all the others? (You can leave, your answer in the form of a sum.) In the \(N \rightarrow \infty\) limit, is the field infinite or finite? In the \(N \rightarrow \infty\) limit, is the force on one of the charges infinite or finite?

Maximum field from a ring ** A charge \(Q\) is distributed uniformly around a thin ring of radius \(b\) that lies in the \(x y\) plane with its center at the origin. Locate the point on the positive \(z\) axis where the electric field is strongest.

Flux through a cube (a) A point charge \(q\) is located at the center of a cube of edge \(d\). What is the value of \(\int \mathbf{E} \cdot d \mathbf{a}\) over one face of the cube? (b) The charge \(q\) is moved to one corner of the cube. Now what is the value of the flux of \(\mathbf{E}\) through each of the faces of the cube? (To make things well defined, treat the charge like a tiny sphere.)

Field from two sheets : Two infinite plane sheets of surface charge, with densities \(3 \sigma_{0}\) and \(-2 \sigma_{0}\), are located a distance \(\ell\) apart, parallel to one another. Discuss the electric field of this system. Now suppose the two planes, instead of being parallel, intersect at right angles. Show what the field is like in each of the four regions into which space is thereby divided.

Oscillating on a line ** Two positive point charges \(Q\) are located at points \((\pm \ell, 0) .\) A particle with positive charge \(q\) and mass \(m\) is initially located midway between them and is then given a tiny kick. If it is constrained to move along the line joining the two charges \(Q\), show that it undergoes simple harmonic motion (for small oscillations), and find the frequency.

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