Electric field above an off-center finite line charge

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SUMMARY

The discussion focuses on calculating the electric field above an off-center finite line charge with a length of 1 meter and a linear charge density of λ = 1×10-16 C/m. The point of interest, P, is located 70 cm above the line charge. Participants utilized Gauss's Law and trigonometric relationships to derive the electric field components Ex and Ey. The final expressions for the electric field's magnitude and direction were determined through vector analysis, emphasizing the importance of correctly defining variables and limits in integrals.

PREREQUISITES
  • Understanding of Gauss's Law and its application in electrostatics
  • Familiarity with trigonometric functions and their use in vector decomposition
  • Knowledge of integral calculus for evaluating electric field components
  • Ability to perform vector analysis to determine magnitude and direction
NEXT STEPS
  • Review the application of Gauss's Law in non-uniform charge distributions
  • Study trigonometric identities and their relevance in vector calculations
  • Explore advanced integral techniques for evaluating electric fields
  • Investigate common pitfalls in vector analysis and integral setup
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, as well as educators seeking to enhance their understanding of electric fields due to finite line charges.

Phezik

Homework Statement


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There is a finite line charge with length L = 1 meter and linear charge density λ = 1*10^-16 C/m. Point P is h = 70cm above the line charge and distance x from the right end of the line charge. The magnitude and direction of the electric field at point P must be found. The final answer should only contain one variable, unknown length x.

Homework Equations


Gauss's Law: ∫E*dA = Q/ε0
Trig Equations: a2+b2=c2, cosθ=adjacent/hypotenuse, sineθ=opposite/hypotenuse
Vector Equations: Magnitude of a vector E with components Ex and Ey, √(Ex2+Ey2)


The Attempt at a Solution


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I set up two integrals to try to solve this problem one for each component of E, the electric field vector. Ex and Ey. Ey was setup using Gauss's law and and using hypotenuse R for the distance times cosθ to find the y competent of the electric field. The integral was then written in terms of height h and distance x and length L for the limits of integration. I then solved the integral to obtain the equation for Ey. I did an identical procedure for Ex except using sinθ to find the x component of the vector.

After finding Ex and Ey I could find E's magnitude using the formula for a vector's magnitude given it's components. I then used cotangent inverse to find the angle (and therefore the direction) of electric field vector E. I added 180° to this angle to find it's angle from the positive x axis.

My theory and execution all seemed correct when I went over it to try to find where I went wrong. I was able to successfully use the same method to find the electric field over a line charge where the point was above the middle of the finite line charge by only finding Ey as the Ex components canceled out. I'm thinking there might be an error in my vector analysis as I only had to deal with one component in the previous problem.
 
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The way the question is written has lured you into a trap.
The x in the diagram is a constant, whereas the dx refers to the length an element of the charge at offset x, a different x.
I suggest changing the element offset to y and using dy for its length.
 
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I tried solving for Ex using your input. I used dy for the integral and used offset x as a constant. I am assuming I should solve for Ey using the same method.
 
Here are my updated attempts to solve for Ey and the magnitude and direction(sorry for the low quality images):
TMo7S8p.jpg

7W9a2Ep.jpg
 
How do you get y(y2+h2) for the integral?
Have you tried differentiating that to check?
 
haruspex said:
How do you get y(y2+h2) for the integral?
Have you tried differentiating that to check?

I redid the integral and got y*h^-1*(y^2+h^2)^(-1/2)
 

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