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A point charge \(q\) is located at the center of a spherical shell of radius \(a\) that has a charge \(-q\) uniformly distributed on its surface. Find the electric field (a) for all points outside the spherical shell and (b) for a point inside the shell a distance \(r\) from the center.

Short Answer

Expert verified
The electric field outside the shell is zero, and the electric field inside the shell a distance \(r\) from the center is \(E=\frac{q}{4\pi\epsilon_0 r^2}\).

Step by step solution

01

Calculate Electric Field Outside the Shell

The total charge enclosed within a Gaussian surface outside the shell is given by \(q_{enc} = q - q = 0\). By Gauss's law, \(E=\frac{q_{enc}}{4\pi\epsilon_0 r^2}\), so the electric field outside the shell is zero.
02

Calculate Electric Field Inside the Shell

The electric field inside the shell is due to the point charge \(q\) at the center. The surface charge (\(-q\)) on the shell does not contribute to the electric field at any point inside the shell. Using Gauss's law, the electric field at a point inside the shell a distance \(r\) from the center should be \(E=\frac{q}{4\pi\epsilon_0 r^2}\).

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Key Concepts

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

Point Charge
When we talk about a "Point Charge", we're referring to an electric charge that is concentrated at a single point in space. In physics, this is often a useful simplification to make calculations easier. Even though in reality charges have a size and distribution, the point charge model helps us understand and calculate electric fields without diving into complex distribution patterns. Consider the point charge as a tiny, indivisible piece of charge that is located at a specific point.
Some essential things about point charges:
  • They help us calculate electric fields when combined with other physics laws.
  • The electric field due to a point charge decreases with the square of the distance from the charge.
  • This means that if you're twice as far from the charge, the field is four times weaker!
In the given exercise, a point charge is placed at the center of our spherical system, and it is this central charge that influences the electric field within the shell.
Gauss's Law
Gauss's Law is a powerful tool in electromagnetism. It relates the electric flux (which is the number of electric field lines crossing a surface) to the charge enclosed by that surface. To put it simply, Gauss's Law states that:\[ \Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q_{enc}}{\epsilon_0} \]Here, \(\Phi\) is the electric flux, \(\mathbf{E}\) is the electric field, and \(q_{enc}\) is the total charge enclosed within a closed surface. \(\epsilon_0\) is the permittivity of free space, a constant that shows how electric fields interact with materials.
In the context of the exercise:
  • For points outside the spherical shell, the enclosed charge is zero, leading to no electric field.
  • For points inside, the field depends only on the central point charge \(q\).
Using Gauss’s Law makes it straightforward to determine that no electric field exists outside the shell when the enclosed net charge is zero.
Spherical Shell
A "Spherical Shell" in physics is often a thin, hollow sphere that might carry electric charge on its surface. It serves as a good example to use when learning about electric fields and forces.
When it comes to electric fields around and inside spherical shells:
  • If you are outside the shell, and the total enclosed charge is zero, the electric field is also zero. This is because the shell's own electric field cancels out when looking from the outside.
  • Inside the shell, the field is unaffected by the charges distributed on the shell but is instead influenced by any internal point charges, like the one in the exercise.
This exercise illustrates perfectly the surprising result from physics that a charge inside a spherical shell only looks influenced by other inside charges, seemingly oblivious to outside influences. This is a consequence of how electric fields inside conductors behave, leading to vital insights in both physics lectures and real-world applications.

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Most popular questions from this chapter

The dome of a Van de Graaff generator receives a charge of \(2.0 \times 10^{-4} \mathrm{C}\). Find the strength of the electric field (a) inside the dome, (b) at the surface of the dome, assuming it has a radius of \(1.0 \mathrm{~m}\), and \((\mathrm{c}) 4.0 \mathrm{~m}\) from the center of the dome. (Hint: See Section \(15.6\) to review properties of conductors in electrostatic equilibrium. Also, use that the points on the surface are outside a spherically symmetric charge distribution; the total charge may be considered to be located at the center of the sphere.)

(a) Sketch the electric field lines around an isolated point charge \(q>0 .\) (b) Sketch the electric field pattern around an isolated negative point charge of magnitude \(-2 a\).

Two small identical conducting spheres are placed with their centers \(0.30 \mathrm{~m}\) apart. One is given a charge of \(12 \times 10^{-9} \mathrm{C}\), the other a charge of \(-18 \times 10^{-9} \mathrm{C}\). (a) Find the electrostatic force exerted on one sphere by the other. (b) The spheres are connected by a conducting wire. Find the electrostatic force between the two after equilibrium is reached.

In the Millikan oil-drop experiment, an atomizer (a sprayer with a fine nozzle) is used to introduce many tiny droplets of oil between two oppositely charged parallel metal plates. Some of the droplets pick up one or more excess electrons. The charge on the plates is adjusted so that the electric force on the excess electrons exactly balances the weight of the droplet. The idea is to look for a droplet that has the smallest electric force and assume it has only one excess electron. This strategy lets the observer measure the charge on the electron. Suppose we are using an electric field of \(3 \times 10^{4} \mathrm{~N} / \mathrm{C}\). The charge on one electron is about \(1.6 \times 10^{-19} \mathrm{C}\). Estimate the radius of an oil drop of density \(858 \mathrm{~kg} / \mathrm{m}^{3}\) for which its weight could be balanced by the electric force of this field on one electron. (Problem 38 is courtesy of E. F. Redish. For more problems of this type, visit www.physics.umd.edu/perg/.)

(a) Sketch the electric field pattern set up by a positively charged hollow sphere. Include regions inside and regions outside the sphere. (b) A conducring cube is given a positive charge. Sketch the electric field pattern both inside and outside the cube.

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