Electric field from a circle arc

Click For Summary
SUMMARY

The discussion focuses on calculating the x-component of the electric field at the origin due to a circular arc with a charge of 3.8 μC and a radius of 1.9 m. The relevant equations include E = kq/r², dq = λ dθ, and λ = Q/(R θ). The user attempted to integrate the expression kq/r² ∫sinθ/θ dθ but received an incorrect answer of 12968.1 N/C. The correct approach involves substituting the values for dq and ds into the expression for dE and integrating from θ = 0 to π/2.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of linear charge density (λ) and its calculation
  • Basic principles of electrostatics and charge distribution
NEXT STEPS
  • Review the derivation of electric fields from continuous charge distributions
  • Practice integration techniques involving trigonometric functions
  • Learn about the application of the Coulomb constant in electric field calculations
  • Explore the concept of linear charge density and its implications in electrostatics
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in solving electrostatic problems related to charge distributions.

nautola
Messages
16
Reaction score
0

Homework Statement


Find the x-component of the electric field
at the origin due to the full arc length
for a charge of 3.8 μC and a radius of
1.9 m. The value of the Coulomb constant
is 8.98755 × 109 N · m2/C2.


Homework Equations


E = kq/r^2
dq = q dθ
λ = Q/ (R θ)


The Attempt at a Solution


I said E = kq/r^2
And so Ex = kq/r2 * sinθ
and dEx = k*dq/r2 * sinθ
and dq = λ ds
ds = R dθ
λ = q/s
so dq = Q/θ dθ

I get kq/r2 ∫sinθ/ θ dθ

But when I input the answer from this integral (from the calculator) the answer is wrong.
 

Attachments

  • 2012-09-12_0155.png
    2012-09-12_0155.png
    4.3 KB · Views: 665
Physics news on Phys.org
What is your answer?
 
I got 12968.1 N/C. Is my method correct?
 
I don't think so.
dq = λ*ds, where
λ= q / s , ds = r*dθ and s = πr/2
Substituute these values in the expression of dE and find the integration taking limits θ = 0 to π/2
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
731
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 79 ·
3
Replies
79
Views
10K
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K