Ring of charge electric potential

Click For Summary

Homework Help Overview

The discussion revolves around calculating the electric potential at a point located 2 meters away from the center of a ring of charge, which lies in the z,y plane. The parameters of the problem include the radius of the ring, the total charge, and the distance from the center to the point of interest.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the integration of the electric potential expression and question the limits of integration. There is an exploration of the variables involved, particularly focusing on the angle θ.

Discussion Status

The conversation is ongoing, with participants clarifying the question and confirming the integration limits. Some guidance has been provided regarding the integration process, but there is no explicit consensus on the next steps.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may impose specific methods or formats for presenting their solutions. There is an emphasis on understanding the setup and the variables involved in the problem.

Panphobia
Messages
435
Reaction score
13

Homework Statement


The ring is in the z,y plane.
D = 2m
R = 4m
Q = 8 * 10^-6 C
X = distance to the edge
λ = dQ/dX
dX = rdθ
dQ = infinitely small charge
dX = infinitely small arc of ring

What is the potential difference at the point D perpendicular to the centre of the ring?

Homework Equations


\int \frac{kdQ}{X}

The Attempt at a Solution



\int \frac{kλrdθ}{(D^2+R^2)^{1/2}}
After this how do I get it in terms of dθ?
 
Last edited:
Physics news on Phys.org
What is the question ? Are you required to find the potential of the ring at a point 2m away from the center ,and lying on the axis of the ring ?
 
Yea I gave the question after all the variables were specified
 
What is the problem ? Just integrate the expression with appropriate limits.The variable of integration is θ .
 
Thats it?
 
Panphobia said:
Thats it?

What does that mean ?
 
So what are we integrating to 0 to 2pi?
 
Yes... the range of θ is 0 to 2pi .
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 25 ·
Replies
25
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 18 ·
Replies
18
Views
1K