Chapter 23: Problem 45
Two charged concentric spherical shells have radii \(10.0 \mathrm{~cm}\) and \(15.0 \mathrm{~cm}\). The charge on the inner shell is \(4.00 \times 10^{-8} \mathrm{C}\), and that on the outer shell is \(2.00 \times 10^{-8} \mathrm{C}\). Find the electric field (a) at \(r=12.0 \mathrm{~cm}\) and (b) at \(r=20.0 \mathrm{~cm}\).
Short Answer
Step by step solution
Understanding the Problem
Electric Field Inside a Conductor
Using Gauss's Law for Inner Region
Calculating Electric Field at r = 12.0 cm
Electric Field Outside Both Shells
Using Gauss's Law for Outer Region
Calculating Electric Field at r = 20.0 cm
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gauss's Law
- \( \Phi_E \) 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 by the surface.
- \( \varepsilon_0 \) is the vacuum permittivity constant.
Spherical Shells
Electrostatics
- Like charges repel and opposite charges attract.
- The electric field describes the force a charge would experience in space.
- Electric field lines start on positive charges and end on negative charges.
- The potential energy is related to the position of charges relative to one another.
Electric Charge
Gaussian Surface
- Spherical for spherically symmetric charges
- Cylindrical for line charges
- Planar for infinite plane charges