Chapter 22: Problem 47
Negative charge \(-Q\) is distributed uniformly over the surface 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 distance (a) \(r > R\) from the center of the shell (outside the shell); (b) \(r < R\) from the center of the shell (inside the shell).
Short Answer
Step by step solution
Understand the Problem Statement
Apply Gauss's Law for Outside the Shell \( (r > R) \)
Calculate the Force Magnitude for \( r > R \)
Determine the Force Direction for \( r > R \)
Apply Gauss's Law for Inside the Shell \( (r < R) \)
Conclude with Force for \( r < R \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coulomb's Law
The formula for Coulomb's Law is:
- \[ F = \frac{k \cdot |q_1| \cdot |q_2|}{r^2} \]
- \( F \) is the magnitude of the force.
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \; \text{Nm}^2/\text{C}^2 \).
- \( q_1 \) and \( q_2 \) are the magnitudes of the two charges.
- \( r \) is the distance between the charges.
Gauss's Law
Formally, Gauss's Law is expressed as:
- \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]
- \( \Phi_E \) is the electric flux through the surface.
- \( \mathbf{E} \) is the electric field.
- \( d\mathbf{A} \) is a differential area on the closed surface.
- \( Q_{\text{enc}} \) is the total charge enclosed within the surface.
- \( \varepsilon_0 \) is the vacuum permittivity, approximately \( 8.854 \times 10^{-12} \; \text{C}^2/\text{Nm}^2 \).
Spherical Shell
Understanding how a spherical shell behaves with charged particles is crucial. Especially when applying laws like Gauss's and Coulomb's. For a point charge outside of the spherical shell,
- The shell's effect is as though all its charge were concentrated at its center.
- This results in an electric field similar to a point charge, calculable through Coulomb's Law.
- Gauss's Law reveals that the internal electric field is zero.
- The charge inside feels no net force from the shell, indicating a unique and important electrostatic property.