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A thin-walled metal spherical shell has radius \(25.0 \mathrm{~cm}\) and charge \(2.00 \times 10^{-7} \mathrm{C}\). Find \(E\) for a point (a) inside the shell, (b) just outside it, and (c) \(3.00 \mathrm{~m}\) from the center.

Short Answer

Expert verified
(a) 0 N/C (b) 2.88 x 10^4 N/C (c) 199.78 N/C

Step by step solution

01

Understanding Electric Field Inside the Shell

Gauss's Law tells us that the electric field inside a conductor in electrostatic equilibrium is zero. With zero net charge enclosed inside the shell, the electric field inside the shell is zero.
02

Electric Field Just Outside the Shell

Just outside the shell, all the charge can be treated as if it were concentrated at the center of the sphere. The electric field at the surface can be found using the formula for the electric field around a point charge: \[ E = \frac{kQ}{r^2} \] where \( k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \), \( Q = 2.00 \times 10^{-7} \, \mathrm{C} \) and \( r = 0.25 \, \mathrm{m} \).
03

Calculating Electric Field Just Outside the Shell

Substitute the values into the formula: \[ E = \frac{(8.99 \times 10^9) \times (2.00 \times 10^{-7})}{(0.25)^2} \] Simplifying this gives \[ E = \frac{1.798 \times 10^3}{0.0625} \approx 2.88 \times 10^4 \, \mathrm{N/C} \].
04

Electric Field at a Point 3.00 m from the Center

For a point outside the shell, the electric field is given by the same formula since the shell behaves as a point charge: \[ E = \frac{kQ}{r^2} \] where now \( r = 3.00 \, \mathrm{m} \).
05

Calculating Electric Field at a Point 3.00 m Away

Substitute the values into the formula: \[ E = \frac{(8.99 \times 10^9) \times (2.00 \times 10^{-7})}{(3.00)^2} \] Simplifying this gives: \[ E = \frac{1.798 \times 10^3}{9} \approx 199.78 \, \mathrm{N/C} \].

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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 vector field that surrounds electric charges. It represents the force a charge would experience if placed in this field. The direction of the electric field is the direction of the force experienced by a positive charge. For a charged spherical shell, understanding the electric field involves concepts from Gauss's Law.

Using Gauss's Law, we can analyze situations where symmetry simplifies calculations. This law helps us determine the electric field at different points relating to a charged object. In the context of a spherical shell:
  • Inside the shell: Gauss's Law states that the electric field (E) is zero because the net charge enclosed by any Gaussian surface inside the conductor is zero.
  • Just outside the shell: The electric field can be calculated assuming all the shell's charge is at its center. The formula \( E = \frac{kQ}{r^2} \) is used, where \( k \) is Coulomb's constant, \( Q \) is the total charge, and \( r \) is the distance from the center.
  • Far from the shell: As the distance increases, the electric field decreases according to the inverse square law, still treatable as if all the charge were a point charge at the center.
Thus, understanding the electric field around a spherical shell involves applying symmetry and using Gauss's law effectively to determine field strength.
Electrostatic Equilibrium
In electrostatic equilibrium, all excess charges in a conductor rest on its surface, and the electric field inside is zero. This happens because any electric field within a conductor would move charges until they are evenly dispersed on the surface, achieving a balanced state.

For a spherical shell:
  • All charges reside on the exterior surface. This is because charges repel each other and move until the repulsive forces are minimized.
  • The electric field inside is zero because the net charge enclosed by any test area inside is zero; hence, there's no net force acting in any direction.
  • The surface charge distribution adjusts itself such that the potential inside remains constant.
This principle explains why the electric field inside a hollow conducting shell in electrostatic equilibrium is zero.
Spherical Shell
A spherical shell is a three-dimensional object characterized by its thickness being negligible compared to its radius, making it essentially a hollow sphere. In problems involving electric fields and charges, a spherical shell acts in a unique way.

Some important points to note for a charged spherical shell include:
  • The net charge is usually distributed uniformly on the surface when the shell is conductive.
  • Gauss's Law simplifies the analysis because you can imagine all charge resides at the center when measuring outside the shell.
  • This model simplifies real-world applications, like modeling hollow conductors in electrostatics.
Hence, the spherical shell is a fundamental part of many problems dealing with electric fields and charge distribution in physics, thanks to its symmetry and simple charge distribution.

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Most popular questions from this chapter

Figure 23-43 shows cross sections through two large, parallel, nonconducting sheets with identical distributions of positive charge with surface charge density \(\sigma=1.77 \times 10^{-22} \mathrm{C} / \mathrm{m}^{2}\). In unit- vector notation, what is \(\vec{E}\) at points (a) above the sheets, (b) between them, and (c) below them?

A uniform charge density of \(500 \mathrm{nC} / \mathrm{m}^{3}\) is distributed throughout a spherical volume of radius \(6.00 \mathrm{~cm} .\) Consider a cubical Gaussian surface with its center at the center of the sphere. What is the electric flux through this cubical surface if its edge length is (a) \(4.00 \mathrm{~cm}\) and (b) \(14.0 \mathrm{~cm} ?\)

A free electron is placed between two large, parallel, nonconducting plates that are horizontal and \(2.3 \mathrm{~cm}\) apart. One plate has a uniform positive charge; the other has an equal amount of uniform negative charge. The force on the electron due to the electric field \(\vec{E}\) between the plates balances the gravitational force on the electron. What are (a) the magnitude of the surface charge density on the plates and (b) the direction (up or down) of \(\vec{E}\) ?

The electric field just above the surface of the charged conducting drum of a photocopying machine has a magnitude \(E\) of \(2.3 \times 10^{5} \mathrm{~N} / \mathrm{C} .\) What is the surface charge density on the drum?

An electron is shot directly toward the center of a large metal plate that has surface charge density \(-2.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) If the initial kinetic energy of the electron is \(1.60 \times 10^{-17} \mathrm{~J}\) and if the electron is to stop (due to electrostatic repulsion from the plate) just as it reaches the plate, how far from the plate must the launch point be?

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