Potential and intensity of magnetic field

AI Thread Summary
The discussion centers on calculating the electric potential and intensity of the electric field at point A due to a uniformly charged quarter-circle wire. Participants confirm that integrating along the wire's length is correct and discuss how to account for changing angles in their calculations. The conversation highlights the use of the relationship dx = R.dθ for integration limits from 0 to π/2. There is also a focus on understanding the gradient of the electric potential and its implications for calculating the electric field. The participants express appreciation for the guidance received, indicating a clearer understanding of the problem.
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Homework Statement


Very thin wire bent into the shape of a quarter-circle with the radius R is uniformly charged with electric charge q.
Calculate the potential and intensity of the electric field at point A, which lies on a line perpendicular to the plane of the semicircle and is passing through the center of its curvature at a distance a from the center. The wire is in the vacuum.

Homework Equations


I am integrating through the length of the conductor, where all the variables seem to be the constants. Is it correct?
How would it look like if I took into account the changing of the angle?

The Attempt at a Solution


see the picture

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this is my first post, greetings to all of you and thanks for the help
Lukas
 

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Welcome to PF.
I am integrating through the length of the conductor, where all the variables seem to be the constants. Is it correct?
Yes.
How would it look like if I took into account the changing of the angle?
You mean if you take θ anticlockwise about the axis, from one end of the wire?
Try it and see: dx = R.dθ and the limits of integration are 0 - π/2 ...
 
Thx a lot for your reply.

Simon Bridge said:
You mean if you take θ anticlockwise about the axis, from one end of the wire?
Try it and see: dx = R.dθ and the limits of integration are 0 - π/2 ...

Yes...mainly "dx = R.dθ" made the trick, thanks again. (the result is the same :) )

Now everything is ready for the second question:
I need to calculate the intensity, I was thinking of using the equation "(vector)E = - grad θ" which gives me the partial derivations by x,y,z.
But θ doesn't contain x,y or z, so I would stay only with constants and therefore 0...what am I missing?
 
Last edited:
##\vec E = -\nabla \phi## ... ;) where ##\phi## is the electric potential. I used ##\theta## for the angle.
How to write equations: https://www.physicsforums.com/help/latexhelp/

You have only worked out the electric potential at a specific point.
##\phi## is not the same everywhere... therefore it has a gradient someplace.
But it may be easier to do the vector calculus for the electric field directly: exploit the symmetry.
 
Simon Bridge said:
##\vec E = -\nabla \phi## ... ;) where ##\phi## is the electric potential. I used ##\theta## for the angle.
How to write equations: https://www.physicsforums.com/help/latexhelp/

I really like the logic behind, it recommends me LibreOffice and its writing of the equations.

Simon Bridge said:
You have only worked out the electric potential at a specific point.
##\phi## is not the same everywhere... therefore it has a gradient someplace.
But it may be easier to do the vector calculus for the electric field directly: exploit the symmetry.

Thx for the guidance, it is clearer now, hope I will do the calculations correctly.
 
ciso112 said:
I really like the logic behind, it recommends me LibreOffice and its writing of the equations.
It's a typsetting markup script called LaTeX - and it's pretty much the academic standard ... you can use it for Libre Office too, though, as you've noticed, there is a limited Tex support already built in.
http://extensions.libreoffice.org/extension-center/texmaths-1

Thx for the guidance, it is clearer now, hope I will do the calculations correctly.
No worries - it can help to explicitly lay out the axes ... I used cartesian with the charges in the positive quadrant of the x-y plane and A on the z axis.
 
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