/*! 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] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 50

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Zero field inside a cylindrical shell * Consider a distribution of charge in the form of a hollow circular cylinder, like a long charged pipe. In the spirit of Problem 1.17, show that the electric field inside the pipe is zero.

Charges on a circular track Suppose three positively charged particles are constrained to move on a fixed circular track. If the charges were all equal, an equilibrium arrangement would obviously be a symmetrical one with the particles spaced \(120^{\circ}\) apart around the circle. Suppose that two of the charges are equal and the equilibrium arrangement is such that these two charges are \(90^{\circ}\) apart rather than \(120^{\circ} .\) What is the relative magnitude of the third charge?

Decreasing energy? A hollow spherical shell with radius \(R\) has charge \(Q\) uniformly distributed over it. The task of Problem \(1.32\) is to show that the energy stored in this system is \(Q^{2} / 8 \pi \epsilon_{0} R\). (You can derive this here if you want, or you can just accept it for the purposes of this problem.) Now imagine taking all of the charge and concentrating it in two point charges \(Q / 2\) located at diametrically opposite positions on the shell. The energy of this new system is \((Q / 2)^{2} / 4 \pi \epsilon_{0}(2 R)=\) \(Q^{2} / 32 \pi \epsilon_{0} R\), which is less than the energy of the uniform spherical shell. Does this make sense? If not, where is the error in this reasoning?

Deriving the energy density ** Consider the electric field of two protons a distance \(b\) apart. According to Eq. (1.53) (which we stated but did not prove), the potential energy of the system ought to be given by $$ \begin{aligned} U &=\frac{\epsilon_{0}}{2} \int \mathbf{E}^{2} d v=\frac{\epsilon_{0}}{2} \int\left(\mathbf{E}_{1}+\mathbf{E}_{2}\right)^{2} d v \\ &=\frac{\epsilon_{0}}{2} \int \mathbf{E}_{1}^{2} d v+\frac{\epsilon_{0}}{2} \int \mathbf{E}_{2}^{2} d v+\epsilon_{0} \int \mathbf{E}_{1} \cdot \mathbf{E}_{2} d v \end{aligned} $$ where \(\mathbf{E}_{1}\) is the field of one particle alone and \(\mathbf{E}_{2}\) that of the other. The first of the three integrals on the right might be called the "electrical self-energy" of one proton; an intrinsic property of the particle, it depends on the proton's size and structure. We have always disregarded it in reckoning the potential energy of a system of charges, on the assumption that it remains constant; the same goes for the second integral. The third integral involves the distance between the charges. Evaluate this integral. This is most easily done if you set it up in spherical polar coordinates with one of the protons at the origin and the other on the polar axis, and perform the integration over \(r\) before the integration over \(\theta\). Thus, by direct calculation, you can show that the third integral has the value \(e^{2} / 4 \pi \epsilon_{0} b\), which we already know to be the work required to bring the two protons in from an infinite distance to positions a distance \(b\) apart. So you will have proved the correctness of Eq. (1.53) for this case, and by invoking superposition you can argue that Eq. (1.53) must then give the energy required to assemble any system of charges.

Find a geometrical arrangement of one proton and two electrons such that the potential energy of the system is exactly zero. How many such arrangements are there with the three particles on the same straight line? You should find that the ratio of two of the distances involved is the golden ratio.
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Solid State Physics

Read Explanation

Quantum Physics

Read Explanation

Rotational Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.