Problem 49 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(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 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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.

Step 5: 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.

Key Concepts

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.