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22.48. Negative charge \(-Q\) is distributed uniformly over the sur- face of a thin spherical insulating shell with radius \(R\) . Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located (a) a distance \(r>R\) from the center of the shell (outside the shell) and (b) a distance \(r

Short Answer

Expert verified
(a) For \(r > R\), the force is \(\frac{k \cdot Q \cdot q}{r^2}\), directed towards the center of the shell. (b) For \(r < R\), the force is zero.

Step by step solution

01

Understanding the Problem for r > R

First, recognize that for a point charge located outside a uniformly charged spherical shell, the shell can be treated as if all its charge were concentrated at the center, according to Gauss's Law.
02

Applying Coulomb's Law for r > R

For a point charge at distance \(r > R\), use Coulomb's Law to calculate the force: \[ F = \frac{k \cdot Q \cdot q}{r^2} \]where \(k\) is Coulomb's constant, \(Q\) is the magnitude of the charge on the shell, and \(q\) is the point charge.
03

Determine Force Direction for r > R

Since the charge on the shell is negative \(-Q\) and the point charge is positive \(q\), the force will be attractive and directed toward the center of the shell.
04

Analysis for r < R Using Gauss's Law

Inside the shell \(r < R\), due to the symmetry of the charge distribution and Gauss's Law, the electric field inside a spherical shell of charge is zero.
05

Conclude Force for r < R

Since the electric field is zero inside the shell, the force exerted on a charge placed inside \(r < R\) is also zero.

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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 for understanding electric fields and charges. It states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. Think of it as a way to simplify complex problems involving electric fields and charges. When dealing with a spherical shell, it helps us conclude that the electric field outside the shell behaves as though all the charge is centrally located.

Gauss's Law is given by the equation \( \oint_{S} \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \), where:
  • \( \mathbf{E} \) is the electric field.
  • \( d\mathbf{A} \) is a differential area on the closed surface \( S \).
  • \( Q_{enc} \) is the total charge enclosed within \( S \).
  • \( \varepsilon_0 \) is the permittivity of free space.

When examining the thin insulating shell in the problem, Gauss's Law assures us that no electric field is present inside the shell \( r < R \). This is because the enclosed charge is zero when considering a Gaussian surface entirely within the shell. Consequently, any charge inside the shell feels no force.
Coulomb's Law
Coulomb's Law is a critical tool for calculating the force between charges. It serves to quantify how much force one charge exerts on another and is pivotal in electrostatics. This force depends on the magnitude of the charges and their separation distance. For our exercise, Coulomb's Law helps calculate the force on a charge outside a charged spherical shell.

The law is expressed as \( F = \frac{k \cdot |Q| \cdot |q|}{r^2} \), where
  • \( F \) is the magnitude of the force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
  • \( Q \) and \( q \) are the magnitudes of the charges.
  • \( r \) is the distance between the charges.

When applying this to a point charge beyond the shell \( r > R \), Coulomb's Law calculates the force directed towards the center, showcasing the attractive interaction due to opposite charges \(-Q\) and \( q \). This formalism allows for precise determination of the force's strength and direction.
Electric Field
The electric field is an essential concept in electrostatics, describing the force per unit charge exerted on a small positive test charge placed in the field. It helps us understand how charged objects influence space around them. The field's presence and characteristics depend on the source charges’ distribution.

For a spherical shell with uniform negative charge distribution, the electric field at any external point behaves as if all the charge is concentrated at the center. Thus, for points \( r > R \), the electric field is given by \( E = \frac{k \cdot Q}{r^2} \). The electric field's direction is radially inward toward the shell, influenced by the shell's negative charge.

Inside the shell \( r < R \), Gauss's Law reveals the electric field vanishes because the shell's symmetric charge distribution results in no net electric field within its cavity. This means a point charge inside feels no force, highlighting how electric fields guide our understanding of charge interactions.

Understanding electric fields is crucial for solving various electrostatic problems, as they offer a comprehensive picture of how charges interact at a distance.

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Most popular questions from this chapter

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.

22.36. A long line carying a uniform linear charge density \(+50.0 \mu C / m\) runs parallel to and 10.0 \(\mathrm{cm}\) from the surface of a large, flat plastic sheet that has a uniform surface charge density of \(-100 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. Find the location of all points where an \(\alpha\) particle would feel no force due to this arrangement of charged objects.

22.23. A hollow, conducting sphere with an outer radius of 0.250 \(\mathrm{m}\) and an inner radius of 0.200 \(\mathrm{m}\) has a uniform surface charge density of \(+6.37 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) A charge of \(-0.500 \mu \mathrm{C}\) is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric fiux through a spherical surface just inside the inner surface of the sphere?

22.54. A Uniformiy Charged Stab. A slab of insulating material has thickness 2\(d\) and is oriented so that its faces are parallel to the \(y z\) -plane and given by the planes \(x=d\) and \(x=-d\) . The \(y\) - and \(z\) -dimensions of the slab are very large compared to \(d\) and may be treated as essentially infinite. The slab has a uniform positive charge density \(\rho\) . (a) Explain why the electric field due to the slab is zero at the center of the slab \((x=0)\) . (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space.

22.1. A flat sheet of paper of area 0.250 \(\mathrm{m}^{2}\) is oriented so that the normal to the sheet is at an angle of \(60^{\circ}\) to a uniform electric field of magnitude 14 \(\mathrm{N} / \mathrm{C}\) (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not?(c) For what angle \(\phi\) between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

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