Electric Potential Field Calculation

Click For Summary
SUMMARY

The discussion focuses on calculating the electric potential field generated by a finite line of charges shaped into two orthogonal segments and a quarter-circle arc. The user initially attempted to apply the superposition principle using three integrals: \(V_1\), \(V_2\), and \(V_3\), but encountered difficulties primarily with the radius calculations, particularly for the segment along the z-axis. The correct approach involves integrating the charge density \(\rho_l\) over the specified segments, ensuring proper limits and angles are applied in the calculations.

PREREQUISITES
  • Understanding of electric potential field equations
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of charge density concepts
  • Basic understanding of superposition in electrostatics
NEXT STEPS
  • Review the derivation of the electric potential field from line charges
  • Practice solving integrals involving charge density in different geometries
  • Learn about the implications of superposition in electrostatic problems
  • Explore the differences between line integrals and surface integrals in electrostatics
USEFUL FOR

Students and professionals in physics or engineering fields, particularly those focusing on electromagnetism and electrostatics, will benefit from this discussion.

nownow
Messages
4
Reaction score
2
Homework Statement
A finite line of charges has been bent into the shape shown in the figure below (2 orthogonal segments and an arc of a circle). Knowing the charge density in the line is given by ρl (C/m) get the electric potential field at the point (0,0,h). h>a
Relevant Equations
electric potential field equation
1655672351756.png

I've already tried to calculate the potential with respect to the 3 segments and then apply superposition (V1+V2+V3). However, I was not very successful. My error I think is in the calculation of the radii, mainly of the line segment that is on the z axis. Can anybody help me? I need some light on this question, please
 
Physics news on Phys.org
nownow said:
Homework Statement:: A finite line of charges has been bent into the shape shown in the figure below (2 orthogonal segments and an arc of a circle). Knowing the charge density in the line is given by ρl (C/m) get the electric potential field at the point (0,0,h). h>a
Relevant Equations:: electric potential field equation

View attachment 303063
I've already tried to calculate the potential with respect to the 3 segments and then apply superposition (V1+V2+V3). However, I was not very successful. My error I think is in the calculation of the radii, mainly of the line segment that is on the z axis. Can anybody help me?
We can help you if you show us what you did. Show us your work and where you think you went wrong. Specifically, what radii are you talking about? There is only one radius and that is associated with the quarter-circle. Your superposition approach is correct.
 
  • Like
Likes   Reactions: bob012345
kuruman said:
We can help you if you show us what you did. Show us your work and where you think you went wrong. Specifically, what radii are you talking about? There is only one radius and that is associated with the quarter-circle.
It's not multiple rays. I expressed myself badly, sorry. However, on to my calculations:

$$V_1 = \int_{0}^{a}\frac{\rho_l}{4\pi \varepsilon_0\sqrt{x^2+h^2} }dx$$
$$V_2 = \int_{0}^{a}\frac{\rho_l}{4\pi \varepsilon_0(h-z) }dz$$
$$V_3 = \int_{0}^{\frac{\pi }{2}}\frac{\rho_la}{4\pi \varepsilon_0\sqrt{a^2+h^2} }d\phi$$

Is what I did correct?
 
what is the angle between the x and y axis?
 
Orodruin said:
what is the angle between the x and y axis?
The angle is ##\frac{\pi }{2}##
 
nownow said:
The angle is ##\frac{\pi }{2}##
So why does your integral specify integration between zero and ##\pi/4##?

Also, editing out an error from your post after getting a reply is not good form.
 
I ended up realizing my error shortly after sending the answer. Also, I didn't edit my answer because of your question. Anyway, I apologize for my mistake and I didn't know about this no edit practice, I'm new here. If you want I can edit the question to what it was before, no problem. I just want to clear my doubt on the question. Again sorry for the edit. Thanks
 
  • Like
Likes   Reactions: PeroK and berkeman
That looks about right. What do you get when you redo the integrals as you have them in post #4 #3? Is your answer the same as you got before?
 
Last edited:
  • Like
Likes   Reactions: Delta2
post #3 setups correctly the integrals. Now the OP has to calculate them correctly.
 
  • Like
Likes   Reactions: kuruman
  • #10
BTW there was another thread, I think from the same OP that the source charge density was portion of a cylindrical surface, so we would have to do a surface integral, I had post to that thread but can't find it anymore, what happened?
 

Similar threads

Replies
2
Views
680
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
1
Views
2K
Replies
10
Views
2K
Replies
1
Views
2K
Replies
2
Views
3K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 22 ·
Replies
22
Views
4K