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Two spherical shells have a common center. A \(-1.6 \times 10^{-6} \mathrm{C}\) charge is spread uniformly over the inner shell, which has a radius of \(0.050 \mathrm{m} . \mathrm{A}+5.1 \times 10^{-6} \mathrm{C}\) charge is spread uniformly over the outer shell, which has a radius of 0.15 \(\mathrm{m}\) . Find the magnitude and direction of the electric field at a distance (measured from the common center) of (a) \(0.20 \mathrm{m},\) (b) \(0.10 \mathrm{m},\) and \((\mathrm{c}) 0.025 \mathrm{m} .\)

Short Answer

Expert verified
(a) 7.88 脳 10鈦 N/C outward, (b) 1.44 脳 10鈦 N/C inward, (c) 0 N/C.

Step by step solution

01

Understand the Problem and Apply Gauss's Law

We have two concentric spherical shells with uniform charge distributions: \(-1.6 \times 10^{-6} \mathrm{C}\) on the inner shell (radius = \(0.050\, \mathrm{m}\)) and \(+5.1 \times 10^{-6} \mathrm{C}\) on the outer shell (radius = \(0.15\, \mathrm{m}\)). We'll use Gauss's Law: \(\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\) to find the electric field at different distances from the common center. Gauss's Law states that the electric flux through a closed surface is proportional to the enclosed charge.
02

Evaluate the Electric Field at 0.20 m (Outside Both Shells)

For a point at \(0.20\, \mathrm{m}\,\), which is outside both shells, the enclosed charge \(Q_{\text{enc}}\) is the sum of the charges of both shells: \(-1.6 \times 10^{-6} \mathrm{C} + 5.1 \times 10^{-6} \mathrm{C} = 3.5 \times 10^{-6} \mathrm{C}\). The electric field \(E\) at this distance is given by \(E = \frac{Q_{\text{enc}}}{4\pi\varepsilon_0 r^2}\), where \(r\) is \(0.20\, \mathrm{m}\). Calculating gives:\(E = \frac{3.5 \times 10^{-6}}{4\pi \times 8.85 \times 10^{-12} \times (0.20)^2} = 7.88 \times 10^{4} \mathrm{N/C}\). The field direction is radially outward due to the net positive charge.
03

Evaluate the Electric Field at 0.10 m (Between the Shells)

At \(0.10\, \mathrm{m}\), the point lies between the two shells. Only the charge on the inner shell contributes to \(Q_{\text{enc}}\), which is \(-1.6 \times 10^{-6} \mathrm{C}\). The electric field is given by:\(E = \frac{-1.6 \times 10^{-6}}{4\pi \times 8.85 \times 10^{-12} \times (0.10)^2} = -1.44 \times 10^{5} \mathrm{N/C}\). The field direction is radially inward due to the negative charge.
04

Evaluate the Electric Field at 0.025 m (Inside the Inner Shell)

At \(0.025\, \mathrm{m}\), the point is inside the inner shell. According to Gauss's Law, no charge is enclosed by a Gaussian surface inside a conductor or shell, so the electric field is zero. Therefore, \(E = 0 \mathrm{N/C}\) at \(0.025\, \mathrm{m}\).

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in electromagnetism. It represents the force per unit charge exerted on a positive test charge at any point in space. Imagine the electric field as an invisible force field that affects charges within its vicinity.
The direction of the electric field is always defined as the direction a positive test charge would move if placed in the field. If the field is created by a positive charge, the direction is outward; for a negative charge, it's inward.
Calculating the electric field involves understanding the charge distribution and the distance from the source charge. Mathematically, for a point charge, the electric field \( E \) is given by:
  • \( E = \frac{kQ}{r^2} \)
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge.
Charge Distribution
Charge distribution refers to how charge is spread out in a given space. In the case of spherical shells, charges are often spread uniformly over the surface.
Uniform charge distribution means every point on the surface has the same charge density. For spherical shells:
  • The shell's symmetry simplifies calculations using Gauss's Law.
  • The total charge, \( Q \), is evenly distributed over the shell.
This uniformity ensures that for points outside the shell, the shell acts like a point charge located at its center. Thus, calculations become straightforward when dealing with spherical symmetry.
Spherical Shells
Spherical shells are three-dimensional structures characterized by their uniform curvature and thickness. They play a crucial role in simplifying electric field calculations.
When dealing with spherical shells:
  • Points outside the shell experience the shell's charge as if it were concentrated at the center.
  • Inside a charged shell, the electric field is zero due to spherical symmetry.
This unique property arises because the internal sections cancel each other out. It's a fantastic reflection of symmetry and Gauss's Law, making it easier to solve complex electrostatic problems. This shell model is often used in introductory physics problems to illustrate fundamental concepts.
Electric Flux
Electric flux represents the number of electric field lines passing through a given surface. It's a vital concept used in Gauss's Law. Think of it as the measure of the 'flow' of the electric field through a surface.
Gauss鈥檚 Law states:
  • The total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.
Mathematically, it is expressed as:
  • \( \Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \)
This principle allows us to calculate electric fields in symmetrical situations easily. By understanding how flux relates to enclosed charge, one can determine the overall field behavior without knowing the detailed field pattern everywhere.

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Most popular questions from this chapter

ssm At a distance \(r_{1}\) from a point charge, the magnitude of the electric field created by the charge is 248 \(\mathrm{N} / \mathrm{C}\) . At a distance \(r_{2}\) from the charge, the ficld has a magnitude of 132 \(\mathrm{N} / \mathrm{C}\) . Find the ratio \(r_{2} / r_{1}\) .

A charge \(Q\) is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: \(\Phi_{1}=+1500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\) \(\Phi_{2}=+2200 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{3}=+4600 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{4}=-1800 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\) \(\Phi_{5}=-3500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) and \(\Phi_{6}=-5400 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} .\) What is \(Q ?\)

Two very small spheres are initially neutral and separated by a distance of 0.50 \(\mathrm{m}\) . Supposc that \(3.0 \times 10^{13}\) electrons are removed from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on each sphere? (b) Is the force attractive or repulsive? Why?

In a vacuum, two particles have charges of \(q_{1}\) and \(q_{2},\) where \(q_{1}=+3.5 \mu \mathrm{C}\) They are separated by a distance of \(0.26 \mathrm{m},\) and particle l experiences an allractive force of 3.4 \(\mathrm{N}\) . What is \(q_{2}\) (magnitude and sign)?

A cube is located with one corner situated at the origin of an x, y, z coordinate system. One of the cube鈥檚 faces lies in the x, y plane, another in the y, z plane, and another in the x, z plane. In other words, the cube is in the first octant of the coordinate system. The edges of the cube are 0.20 m long. A uniform electric field is parallel to the x, y plane and points in the direction of the y axis. The magnitude of the field is 1500 N/C. (a) Using the outward normal for each face of the cube, find the electric flux through each of the six faces. (b) Add the six values obtained in part (a) to show that the electric flux through the cubical surface is zero, as Gauss鈥 law predicts, since there is no net charge within the cube.

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