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Chapter 22: Problem 12

22.12. (a) In a certain region of space, the volume charge density \(\rho\) has a uniform positive value. Can \(\overrightarrow{\boldsymbol{E}}\) be uniform in this region? Explain. (b) Suppose that in this region of uniform positive \(\rho\) there is a "bubble" within which \(\rho=0 .\) Can \(\vec{E}\) be uniform within this bubble? Explain.

Short Answer

Expert verified
(a) No, because a uniform electric field cannot satisfy the positive divergence required by Gauss's Law for a region with uniform positive \( \rho \). (b) Yes, because inside the bubble, the charge density is zero, allowing \( \mathbf{E} \) to be uniform.

Step by step solution

01

Understanding Volume Charge Density

Volume charge density \( \rho \) represents the amount of charge per unit volume. A uniform positive \( \rho \) indicates that the charge is evenly distributed throughout the region.
02

Applying Gauss's Law

By Gauss's Law, \( abla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} \). Since \( \rho \) is uniform and positive, the electric field \( \mathbf{E} \) must diverge from every point, meaning \( \mathbf{E} \) cannot be uniform because a uniform electric field would have zero divergence.
03

Analyzing the Bubble

Inside the bubble where \( \rho = 0 \), Gauss's Law gives \( abla \cdot \mathbf{E} = 0 \). This indicates that \( \mathbf{E} \) can be uniform because the divergence of \( \mathbf{E} \) is zero, allowing a constant electric field inside the bubble.

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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 is a fundamental principle in electromagnetism that connects the electric flux through a closed surface to the charge enclosed by that surface. Its mathematical expression is: \[ abla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} \] where \( abla \cdot \mathbf{E} \) is the divergence of the electric field \( \mathbf{E} \), \( \rho \) is the volume charge density, and \( \varepsilon_0 \) is the permittivity of free space. Key points to understand:
  • Flux through a surface: Gauss's Law focuses on how the electric field lines pass through a closed surface. This is directly related to the amount of charge inside the surface.
  • Symmetry in charge distribution: Gauss's Law is especially helpful in cases where symmetry makes the problem simpler, like spherical or cylindrical charge distributions.
Thus, Gauss鈥檚 Law helps us understand that a uniform electric field cannot exist in a region with non-zero, uniform charge density since such a field would imply zero divergence, contradicting the law.
Volume charge density
Volume charge density \( \rho \) is a measure of electric charge per unit volume of space, typically expressed in coulombs per cubic meter (C/m鲁). Understanding \( \rho \) is crucial because it helps in determining the arrangement and magnitude of charges across a region. Important aspects of volume charge density:
  • Uniform charge distribution: When \( \rho \) is constant throughout a space, it indicates a uniform charge distribution. In such cases, calculating the electric field becomes more manageable using symmetry.
  • Impact on the electric field: The presence and distribution of volume charge density affect the calculation of electric fields using both Gauss's Law and the gradient of potentials.
In example exercises, when \( \rho \) is uniformly positive across a region, it signifies that the electric field originates and diverges from each point, affecting the field's uniformity.
Divergence of electric field
Divergence of the electric field \( abla \cdot \mathbf{E} \) quantifies the electric field's tendency to diverge or converge at a point. It is a vital concept to understand because it helps in analyzing how fields behave over space. Points to consider about electric field divergence:
  • Divergence and source charge: A non-zero divergence in an electric field indicates the presence of source charges, where field lines originate or terminate.
  • Zero divergence conditions: When the divergence of \( \mathbf{E} \) is zero, it usually implies a uniform field in regions where there are no charges, as in a bubble with \( \rho = 0 \). Here, the field can be steady and consistent without "leaking" out or converging.
In scenarios involving a uniform positive volume charge density, like in the exercise, the electric field cannot be uniform due to its inevitable divergence at various points in charged space. However, in regions without charge, such as a bubble with \( \rho = 0 \), the field can indeed remain uniform.

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Most popular questions from this chapter

22.3 You measure an electric field of \(1.25 \times 10^{6} \mathrm{N} / \mathrm{C}\) at a distance of 0.150 \(\mathrm{m}\) from a point charge. (a) What is the electric flux through a sphere at that distance from the charge? (b) What is the magnitude of the charge?

22.15. A point charge of \(+5.00 \mu C\) is located on the \(x\) -axis at \(x=4.00 \mathrm{m},\) next to a spherical surface of radius 3.00 \(\mathrm{m}\) centered at the origin. (a) Calculate the magnitude of the electric field at \(x=3.00 \mathrm{m} .\) (b) Calculate the magnitude of the electric field at \(x=-3.00 \mathrm{m} .\) (c) According to Gauss's law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e., at \(x=3.00 \mathrm{m}\) ) than on the far side (at \(x=-3.00 \mathrm{m}\) ). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)? Explain. A sketch will help.

22\. 14. Electric Fields in an Atom. 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 \(7.4 \times 10^{-15} \mathrm{m}\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{m} ?(\mathrm{c})\) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

22.16. A solid metal sphere with radius 0.450 \(\mathrm{m}\) carries a net charge of 0.250 \(\mathrm{nC}\) . Find the magnitude of the electric field (a) at a point 0.100 \(\mathrm{m}\) outside the surface of the sphere and \((\mathrm{b})\) at a point inside the sphere, 0.100 \(\mathrm{m}\) below the surface.

22.25. The electric field at a distance of 0.145 \(\mathrm{m}\) from the surface of a solid insulating sphere with radius 0.355 \(\mathrm{m}\) is 1750 \(\mathrm{N} / \mathrm{C}\) . (a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? ( 6 ) Calculate the electric field inside the sphere at a distance of 0.200 \(\mathrm{m}\) from the center.
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