Electric field and potential at the origin

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SUMMARY

The discussion focuses on calculating the electric field vector and potential at the origin (0,0) due to a longitudinal charge on a half-circle with radius 'a'. The primary tool referenced is Coulomb's law, which is used to derive the electric field components. The participants highlight the complexity of integrating the electric field to find potential, noting that an alternative method using the formula V = Q'/4πε₀ simplifies the process. The consensus is that while both methods can yield the same result, the electric field approach is more cumbersome and less efficient for this problem.

PREREQUISITES
  • Coulomb's law for electric fields
  • Integration techniques in calculus
  • Understanding of electric potential and its relation to electric fields
  • Knowledge of reference points in electrostatics
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  • Study the derivation of electric potential from electric fields using integration
  • Learn about the concept of reference points in electrostatics
  • Explore alternative methods for calculating electric potential, such as using charge distributions
  • Investigate the implications of using different coordinate systems in electrostatics problems
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Students studying electromagnetism, physics educators, and anyone involved in solving electrostatics problems, particularly those dealing with electric fields and potentials.

cdummie
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Homework Statement


I have to find a electric field vector and a potential of the point in the origin of the x0y coordinate system (0,0) due to a longitudinal charge placed on a half circle as shown in the picture, with radius a.

Homework Equations


Coulombs law.

The Attempt at a Solution


слика.png


Using Coulombs law:
Кулонов закон.png


Vector E has two components:
разлагање вектора.png


Which means E=Ey.

Since dQ=Q'dl=Q'adθ

ипсилон компонента.png


If someone notices any mistakes up here, please point it out, since i haven't noticed any.

Now, for potential:

потенцијал.png


dl=dy because E has only y component, now when i integrate this i have V=∞, which seems wrong to me, so if someone knows how to solve this please post solution here.
 
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Why do you integrate from a to infinity?
Also, E will depend on y. There is an easier way to calculate the potential, without electric fields.
 
mfb said:
Why do you integrate from a to infinity?
Also, E will depend on y. There is an easier way to calculate the potential, without electric fields.

Well i thought, since distance from point whose potential i want to know and charge is a i put first limit a, and since reference point is at infinity (i think i haven't pointed that out in the statement). I think i know what do you mean by easier way its probably dV=dQ/4πξ0a and then since dQ=Q'dl when i integrate it i get V=Q'/4ξ0. Is this correct? If it is then i should get the same thing doing it the way i did up there, but obviously i won't get the same thing as here.
 
Yes that is correct.

cdummie said:
Well i thought, since distance from point whose potential i want to know and charge is a i put first limit a, and since reference point is at infinity
The approach with the electric field would need a path from the reference point to the position where you want to know the potential. And you would have to calculate the electric field everywhere along this path. The second part is quite messy.
 
mfb said:
Yes that is correct.

The approach with the electric field would need a path from the reference point to the position where you want to know the potential. And you would have to calculate the electric field everywhere along this path. The second part is quite messy.
That means that second approach is better for finding potential (i mean it's easier), I've done it that that way, but i thought if i solve it using electric field it could confirm that i did it correctly since the second approach seemed simple and i thought, if it's that simple, then probably i did something wrong. Anyway, thank you for help. :)
 
Solving it via electric fields is much more work. It would be a cross-check, but too tricky for such a homework problem.
 
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