Electric field above an off-center finite line charge

In summary, the problem involves finding the magnitude and direction of the electric field at point P, which is 70cm above a finite line charge with length L = 1m and linear charge density λ = 1*10^-16 C/m. Using Gauss's Law, trigonometric equations, and vector equations, two integrals were set up for the components of the electric field vector, Ex and Ey. After solving the integrals, the magnitude of E was found using the formula for a vector's magnitude and the angle of E was found using cotangent inverse. However, upon further analysis, it was discovered that the offset x in the diagram is a constant, while the dx in the integral refers to the
  • #1
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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  • #2
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.
 
  • #3
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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.
 
  • #4
Here are my updated attempts to solve for Ey and the magnitude and direction(sorry for the low quality images):
TMo7S8p.jpg

7W9a2Ep.jpg
 
  • #5
How do you get y(y2+h2) for the integral?
Have you tried differentiating that to check?
 
  • #6
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)
 

1. What is an electric field?

An electric field is a physical field that surrounds any electrically charged object. It is responsible for the electric force that acts on other charged objects within its range.

2. How is an electric field above an off-center finite line charge different from a regular electric field?

An off-center finite line charge creates an uneven distribution of charge, resulting in a more complex electric field. In contrast, a regular electric field is created by a point charge or a evenly distributed line charge.

3. How is the electric field strength calculated above an off-center finite line charge?

The electric field strength above an off-center finite line charge can be calculated using Coulomb's Law, which states that the electric field strength is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.

4. What factors affect the electric field strength above an off-center finite line charge?

The electric field strength above an off-center finite line charge is affected by the magnitude and distribution of the charge, as well as the distance from the charge. Additionally, the medium in which the charge is located can also impact the electric field strength.

5. How can the direction of the electric field above an off-center finite line charge be determined?

The direction of the electric field can be determined by the orientation of the charged object. The electric field lines always point away from positive charges and towards negative charges. Additionally, the electric field lines are perpendicular to the surface of the charged object at any given point.

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