Work done by electric field of an irregularly shaped conductor

[MP1021]

TL;DR Summary

Why does the charge distribution on an irregualry shaped conductor not result in different work done depending on where an attracted test charge ends up?

I am trying to wrap my head around something and would be grateful for some insight. Specifically, why the different electric fields along different paths to different areas of an irregularly shaped conductor don't impart different energies to inflowing particles.

Say we have a negatively biased conductor with regions of lower and higher surface curvature, so that the electric field will be more intense near sharper (protruding) points as compared to flatter regions of the conductor. A nearby postive test charge, starting from a region identified as at zero electric potential, and with no initial kinetic energy, will be accelerated towards the conductor and will ultimately strike it with an energy entirely determined by the potential of the conductor (and the test particle's charge), yes?

I get why, the electric force being conservative, the path taken to the same point on the conductor won't affect the test charge's final energy, but I can't make sense of why the impacting location won't affect it (just the conductor's potential). It makes sense if I think of it like gravitational potential energy- this being equivalent to two masses reaching the same final speed after travelling the same vertical distances on different slopes if there were no friction or air resistance. But when I think of it from the perspective of the charge distribution, it seems like the charge coming in from the right on the picture below should be more strongly attracted at all times along its path (as compared the particle on the left) and ultimately be moving faster before the impact.

Do I understand the situation correctly? Thanks for taking the time to read this :)

 

[andrewkirk]

I think the problem lies in an incorrect hidden assumption the OP makes: that the ellipsoid shown is a surface of constant potential, so that the two charges start with the same potential.

The electric field outside the conductor is much stronger near the point (on the right) than on the left. So surfaces of equipotential will be much closer to the conductor on the right than on the left.

Put differently, the conductor will lie towards the right end of such surfaces, not approximately centred as shown above.

The work done is the integral of force over distance. A charge coming in from the right end of an equipotential surface will have a greater force but a smaller distance, compared to one coming in from the left. By construction of the surface, these two effects exactly cancel out, so that the work done in moving a particle to the surface from the right is the same as moving one from the left.

 

[Baluncore]

TL;DR Summary: Why does the charge distribution on an irregualry shaped conductor not result in different work done depending on where an attracted test charge ends up?

... , the path taken to the same point on the conductor won't affect the test charge's final energy, but I can't make sense of why the impacting location won't affect it (just the conductor's potential).

Assuming the two surfaces are equipotentials.
The work done is force times distance.
Where the surfaces are closer, the voltage gradient is steeper, so the force is greater.
The product of force by distance remains constant.
The flight time for different paths is different.

The final velocity, is compensated in the same way.
More distance has more time to accelerate with less force due to lower field gradient.

 

[MP1021]

Thanks for your responses andrewkirk and Baluncore. You've both identified that there is a tradeoff between field gradient and distance, with the result of an increase in one compensating for the decrease of the other, which does make sense to me on the scale of the cartoon I drew. It was still counterintuitive to me for smaller, sharper features, which I imagined would have drastically increased field strengths (but only a minute change in distance). But, after playing around with different protrusion sizes in FEMM for a bit, I saw that your explanation does make sense at smaller scales as well. My intuition about the extent of the impact of such protrusions on the electric field was totally off.

Thanks so much for your help!