Solution to Griffiths introduction to EM problem 2.26

In summary, a student asks for the solution to a complicated integral in problem 2.26 of Griffiths Introduction to Electrodynamics. They mention using the same textbook and covering chapter 2 in lectures, but also remind the student that they won't do their homework for them. The student then shares the problem of finding the potential difference between two points on a conical surface with a uniform surface charge. They thank the person for their help and mention a previous post asking for the solution to a specific problem and another asking about the chapters covered on a test.
  • #1
question2004
8
0
Hi, there,

Could you please show me the solution to Griffiths Introduction to Electrodynamics, problem 2.26? The integral is so complicated. Thanks
 
Physics news on Phys.org
  • #2
Cool! We're using the same textbook. Yeah, my next assignment will probably have problems from that section as well, since we're covering chapter 2 in the lectures right now. But, no one's going to do your homework for you. Show what you've got so far, and they'll help you out with corrections/point you in the right direction.
 
  • #3
You might need to post the problem, since many you could help you may not have that text.
 
  • #4
The problem is

A conical surface (an empty ice-cream cone) carries a uniform surface charge <sigma>. The height of the cone is h, and the radius of the top is R. Find the potential difference between points a(the vertex) and b (the center of the top)

Thanks a lot.
 
  • #5
It could be worse!

I remember one post that said, in it's entirety, "What's the solution to the fourth problem on today's homework."

And another "What chapters will tomorrows test cover?"
 

1. What is problem 2.26 in Griffiths introduction to EM?

Problem 2.26 in Griffiths introduction to EM is a question that asks you to find the electric field at a point in space due to a charged ring with a non-uniform charge distribution.

2. What is the solution to problem 2.26?

The solution to problem 2.26 involves using the principle of superposition and integrating the electric field contributions from infinitesimal rings along the circumference of the charged ring. This results in an expression for the electric field at the given point in terms of the charge distribution and the distance from the center of the ring.

3. What are the key concepts needed to solve problem 2.26?

To solve problem 2.26, one needs to understand the concept of electric field, Coulomb's law, and the principle of superposition. It is also helpful to have a strong understanding of integration and how to apply it in the context of electric field calculations.

4. Are there any tips or tricks for solving problem 2.26?

One way to simplify the calculation for problem 2.26 is to choose a coordinate system where the point of interest lies on the z-axis and the charged ring lies on the xy-plane. This way, the electric field contributions from the rings can be easily integrated using polar coordinates.

5. Can the solution to problem 2.26 be applied to other similar problems?

Yes, the solution to problem 2.26 can be applied to other similar problems involving non-uniformly charged rings or disks. However, the charge distribution and geometry of the object may need to be adjusted in the calculation. It is important to understand the underlying principles and adapt them accordingly for different scenarios.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
4K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
935
  • STEM Academic Advising
Replies
11
Views
1K
  • Science and Math Textbooks
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
754
  • Introductory Physics Homework Help
Replies
8
Views
4K
  • Advanced Physics Homework Help
Replies
1
Views
899
Back
Top