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Chapter 1: Problem 27

\(\mathrm{~ U n i f o r m ~ f i e l d ~ i n ~ a ~ c a v i t y ~ \cjkstart ? ? ? ? ? ?}\) A sphere has radius \(R_{1}\) and uniform volume charge density \(\rho . \mathrm{A}\) spherical cavity with radius \(R_{2}\) is carved out at an arbitrary location inside the larger sphere. Show that the electric field inside the cavity is uniform (in both magnitude and direction). Hint: Find a vector expression for the field in the interior of a charged sphere, and then use superposition. What are the analogous statements for the lower-dimensional analogs with cylinders and slabs? Are the statements still true?

Short Answer

Expert verified
The electric field inside the cavity of the sphere and cylinder is uniform and zero in magnitude. For the slab, it's not uniform; it is zero at the center but changes as we move away.

Step by step solution

01

Determine the Electric Field of Charged Sphere

Start by considering a sphere of radius R and uniform volume charge density \(\rho\). The charge enclosed in a spherical Gaussian surface of radius r is \(\frac{4}{3}\pi r^{3}\rho\). According to Gauss's law, the electric field at this point is \(E=\frac{1}{4\pi \epsilon_{0}}\frac{Q_{enc}}{r^{2}}=\frac{1}{4\pi \epsilon_{0}}\frac{4}{3}\pi r^{3}\rho / r^{2} = \frac{1}{3\epsilon_{0}} \rho r\), which is radially outward. Thus, the electric field in vector terms is \(\vec{E}=\frac{1}{3\epsilon_{0}} \rho r \hat{r}\).
02

Calculate the Electric Field inside the Cavity

The electric field inside the cavity is the superposition of the electric field of the charged sphere and the field of the negative charged sphere that forms the cavity. The negative charged sphere has exactly the same properties as the charged sphere, but the charge density will be \(-\rho\). The electric field of the negative charged sphere in vector form is \(\vec{E}=-\frac{1}{3\epsilon_{0}} \rho r \hat{r}\). Thus, when you add these two electric fields, the result is zero. This means that the electric field inside the cavity is uniform and of zero magnitude, meaning it is zero everywhere inside the cavity.
03

Extend the Discussion to Lower-Dimensional Analogs

In the case of a cylinder, a similar analysis would show that the electric field inside the cavity is also zero. However, for a slab (i.e., a three-dimensional object with one dimension significantly smaller than the other two), only the center of the cavity would have zero electric field. As one moves away from the center, the symmetry breaks down and there is an electric field in the cavity. Thus, the statement does not hold for the slab.

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Key Concepts

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

Uniform Charge Density
Uniform charge density is an important concept when dealing with charged objects. It represents a consistent distribution of charge over a given volume. For example, in a sphere of radius \( R_1 \) with a uniform volume charge density \( \rho \), the charge is evenly spread throughout the sphere's material, leading to a predictable electric field. This uniformity simplifies calculations of electric fields, as you don't need to consider variations in charge at different points. It's crucial to know that while the charge is evenly distributed, the electric field generated depends on the position within or outside the sphere.
Gaussian Surface
A Gaussian surface is an imaginary closed surface used in Gauss's law to calculate electric fields. When applying Gauss's law, you choose a surface that simplifies calculations, often a sphere, cylinder, or plane, depending on the symmetry of the charge distribution. In the case of a uniformly charged sphere, the electric field at any point outside is calculated using a spherical Gaussian surface. The choice of a spherical surface simplifies the calculation because it matches the symmetry of the sphere itself. The symmetry ensures that the electric field is constant over the surface, allowing for straightforward integration to find the total electric field.
Superposition Principle
The superposition principle is a cornerstone of understanding electric fields and charges. It states that the total electric field caused by multiple charges is the vector sum of the fields due to each individual charge. This principle is incredibly useful when dealing with complex configurations, such as a charged sphere with a cavity inside. By considering the sphere as composed of two separate charge distributions, one positive and one negative (the cavity), you can calculate the total electric field by adding the fields from both sources. This approach simplifies problems and makes it easier to visualize how fields interact in space.
Cavity in a Charged Sphere
Carving a cavity within a charged sphere introduces complexity, but understanding it is made easier by using the superposition principle. The cavity represents a region with no charge and should ideally have a uniform electric field. By treating the cavity as a negative charge density \( -\rho \) within the positive sphere, you can analyze the electric field effects as superimposed fields. Ultimately, this shows that the electric field within the cavity is uniform and zero, reaffirming that the fields from the positive and negative regions cancel each other out perfectly, preventing any non-zero electric field within the cavity itself. However, this conclusion does not always hold for all geometries, such as slabs, where field symmetry breaks down.

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Most popular questions from this chapter

Field at the end of a cylinder (a) Consider a half-infinite hollow cylindrical shell (that is, one that extends to infinity in one direction) with radius \(R\) and uniform surface charge density \(\sigma .\) What is the electric field at the midpoint of the end face? (b) Use your result to determine the field at the midpoint of a half-infinite solid cylinder with radius \(R\) and uniform volume charge density \(\rho\), which can be considered to be built up from many cylindrical shells.

Fields at the surfaces Consider the electric field at a point on the surface of (a) a sphere with radius \(R\), (b) a cylinder with radius \(R\) whose length is infinite, and (c) a slab with thickness \(2 R\) whose other two dimensions are infinite. All of the objects have the same volume charge density \(\rho\). Compare the fields in the three cases, and explain physically why the sizes take the order they do.

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 \(\rho R / 4 \epsilon_{0}\)

Zero field in a sphere In Fig. \(1.51\) a sphere with radius \(R\) is centered at the origin, an, infinite cylinder with radius \(R\) has its axis along the \(z\) axis, and an infinite slab with thickness \(2 R\) lies between the planes \(z=-R\) and \(z=R\). The uniform volume densities of these objects are \(\rho_{1}, \rho_{2}\) and \(\rho_{3}\), respectively. The objects are superposed on top of each other; the densities add where the objects overlap. How should the three densities be related so that the electric field is zero everywhere throughout the volume of the sphere? Hint: Find a vector expression for the field inside each object, and then use superposition.

Gravity vs. electricity (a) In the domain of elementary particles, a natural unit of mass is the mass of a nucleon, that is, a proton or a neutron, the basic massive building blocks of ordinary matter. Given the nucleon mass as \(1.67 \cdot 10^{-27} \mathrm{~kg}\) and the gravitational constant G as \(6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \mathrm{s}^{2}\right)\), compare the gravitational attraction of two protons with their electrostatic repulsion. This shows why we call gravitation a very weak force. (b) The distance between the two protons in the helium nucleus could be at one instant as much as \(10^{-15} \mathrm{~m}\). How large is the force of electrical repulsion between two protons at that distance? Express it in newtons, and in pounds. Even stronger is the nuclear force that acts between any pair of hadrons (including neutrons and protons) when they are that close together.
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