# Electric field and potential at the origin

• cdummie
In summary, the conversation discusses finding an electric field vector and potential at the origin due to a longitudinal charge on a half circle. Coulomb's law is used and the electric field has two components, with the potential being calculated using the integration of dl=dy. However, this yields an infinite result and a simpler method using dV=dQ/4πξ0a is suggested. The conversation concludes that this second approach is easier and more suitable for this problem.
cdummie

## 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.

Coulombs law.

## The Attempt at a Solution

Using Coulombs law:

Vector E has two components:

Which means E=Ey.

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

Now, for potential:

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.

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.

cdummie

## What is an electric field and potential at the origin?

An electric field at the origin is the force experienced by a charged particle placed at that point. The potential at the origin is the amount of work needed to bring a unit positive charge from infinity to that point.

## How is the electric field and potential at the origin calculated?

The electric field at the origin is calculated using Coulomb's law, which states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The potential at the origin is calculated by taking the negative integral of the electric field with respect to distance.

## What factors affect the electric field and potential at the origin?

The electric field and potential at the origin are affected by the magnitude and sign of the charges involved, as well as the distance between them. Additionally, the presence of other charges in the surrounding environment can also influence the electric field and potential at the origin.

## What is the relationship between electric field and potential at the origin?

The electric field and potential at the origin are related by the equation E = -∇V, where E is the electric field, V is the potential, and ∇ represents the gradient operator. This means that the electric field is equal to the negative of the potential gradient at that point.

## Why is the electric field zero at the origin?

The electric field is zero at the origin because the force between two point charges is inversely proportional to the square of the distance between them. At the origin, the distance between the two charges is zero, resulting in an infinite force. In order for the force to be finite, the electric field must be zero.

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