What is the Electric Potential of a Uniformly Charged Bent Wire?

Click For Summary
SUMMARY

The discussion centers on calculating the electric potential of a uniformly charged bent wire with a linear charge density of λ=6.22×10-9 C/m. The potential is derived using the formula V = k ∫(dq/r), where 'k' is the Coulomb's constant. Participants confirm the integration steps for the left line, circle, and right line segments, while addressing the correct interpretation of 'r' as the distance from the charge to the origin. The final result for the electric potential is 298 V.

PREREQUISITES
  • Understanding of electric potential and Coulomb's law
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of linear charge density concepts
  • Basic proficiency in using LaTeX for mathematical expressions
NEXT STEPS
  • Study the application of Coulomb's law in different geometries
  • Learn advanced integration techniques for complex shapes
  • Explore the use of LaTeX for formatting mathematical equations
  • Investigate the impact of charge distribution on electric potential
USEFUL FOR

Students in physics or electrical engineering, educators teaching electromagnetism, and anyone interested in understanding electric potential calculations for charged objects.

Leeoku
Messages
18
Reaction score
0

Homework Statement


. A wire of finite length that has a uniform linear charge density
λ=6.22×10-9 C/m is bent into the shape shown below.
[PLAIN]http://lulzimg.com/i23/7498af.jpg
Answer: 2.98e+02 V

Homework Equations


V = k integral (dq/r)

The Attempt at a Solution


So i think my integrals are right not sure. I have to split into line and circle.
Left line: K (integral -3r->-R) lambda dx/x
Circle: K (integral -pi->0) lambda d theta
Right line k(Integral R->3R) lambda dx/x

So when i integrate and plug stuff in i get
K*Lambda (ln[-R]-ln[-3r]+pi+ln[3R]-ln[R])
im not sure how to simplify the terms with ln R

side note: How can i properly write stuff out on forums? =S
 
Last edited by a moderator:
Physics news on Phys.org
Leeoku said:
So i think my integrals are right not sure. I have to split into line and circle.
Left line: K (integral -3r->-R) lambda dx/x

This isn't right. In V = k integral (dq/r), r is the distance between the charge and the origin, and can't be negative.

Circle: K (integral -pi->0) lambda d theta
Right line k(Integral R->3R) lambda dx/x

These are right.

So when i integrate and plug stuff in i get
K*Lambda (ln[-R]-ln[-3r]+pi+ln[3R]-ln[R])
im not sure how to simplify the terms with ln R

You can use the identity ln(a)+ln(b)=ln(a*b), or equivalently, ln(a)-ln(b)=ln(a/b).
side note: How can i properly write stuff out on forums? =S

The usual way is to use LaTex. See here for a tutorial: http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/

For most purposes, you can just search up and use Unicode symbols along with the exponent and underscore tags. There are text symbols for all the Greek characters, for some fractions, for the integral sign, and probably for a lot more things I don't know about.
 
i don't understand what to integrate from the left side. How am i supposed to express it with relation to the origin or do i just take the right integral and multiply by 2?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
64
Views
6K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 18 ·
Replies
18
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K