Total force acting on charged particle by charged Wire?

Question

A charged particle Q is placed at distance R away from a semi-infinite charged wire with charge per unit length of σ.

The graph looks something like this:

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Q

The charge is placed at distance R away from the wire, directly under the beginning of the wire


What is the total force acting on the particle?
Electrostatic forces are integrable, but can you derive the answer without actual integration?

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This problem if done 'properly' has a certain flavor of re-normalization to it.

Attempt to calculate potential of the charge with respect to infinity results in divergent integral, but this infinty can be avoided, if potential is calculated with respect to the original position.

Vertical component of the force is (per Gauss theorem & symmetries)
Fy = 2/2 kσ = kσ

Horizontal component is best computed as differential of potential:
Fx = d/dx P(x) = d/dx ∫kσdx /r
The trick is that instead of varying the position of the partice by dx to the right, we can pull the string by distance dx to the left. The increase in potential will be dP = kσ dx due to newly added charge σ dx on the tip of the string, and consequently
Fx = kσ dx/dx = kσ = Fy.

The funny thing is that integral ∫kσdx /r diverges, but the semi-infinite tail is re-normalizable, so that observable potentials and forces are all finite. This solution, I admit, must sound horrible to any mathematician.

Answer 1

Well, doing it with integration, the infinite wire produces a force on the particle proportional to 2/R, while the semi-infinite wire produces a force proportional to (√2)/R.

The angle of the net force exerted by a semi-infinite wire would be the same regardless of R, and it's at 45°, per the calculus work. So that would explain the factors 2 and √2. Now, why is it at 45°? Hmm...

Okay, to the charge particle, the semi-infinite wire looks the same as a quarter circle of uniform charge distribution (easy trig shows this), so therefore, for symmetry reasons, the net force can be expected to be at 45°. Since the apparent charge density of this quarter circle is inversely proportional to R, the force equation would follow. Dah?

Answer 2

This is an interesting challenge. It's clear to me that the vertical component of the force on Q will be just half the force it would experience from an infinite wire, but the horizontal component will be trickier.

My instinct tells me that the angle of the total force vector on Q will be independent of R because a semi-infinite wire will "look the same" at all scales ... I'll have to check whether that instinct is correct.

I've been trying to come up with a clever way to superpose two semi-infinite wires to come up with some insight, but so far I've got nothing productive. I'll keep thinking...