Potential difference between two spherical conductors

[alanf]

I'm working my way through MIT 8.02x on EdX (an archived course, so it's a bit lonely in there right now!). The problem statement:

Two spherical conductors, A and B, are placed in vacuum. A has a radius rA=25 cm and B of rB=35 cm. The distance between the centers of the two spheres is d=225 cm. A has a potential of VA=100 Volt and B has a potential of VB= -25 Volt. An electron is released with zero speed from B. What will its speed be as it reaches A?

I got the approved answer simply by assuming the change in energy equals the charge on the electron times the potential difference. But is it really that simple? Each sphere should have uneven charge distribution because of induction. The field on the surface of sphere A closest to sphere B should be stronger than on the other side of sphere A, and the same applies to sphere B. Doesn't that affect the analysis?

 

[collinsmark]

I'm working my way through MIT 8.02x on EdX (an archived course, so it's a bit lonely in there right now!). The problem statement:

Two spherical conductors, A and B, are placed in vacuum. A has a radius rA=25 cm and B of rB=35 cm. The distance between the centers of the two spheres is d=225 cm. A has a potential of VA=100 Volt and B has a potential of VB= -25 Volt. An electron is released with zero speed from B. What will its speed be as it reaches A?

I got the approved answer simply by assuming the change in energy equals the charge on the electron times the potential difference.

That sounds right to me.

But is it really that simple?

Yes, the electrical potential is path independent.

There are some assumptions such as the enclosure was evacuated (by that I mean no air), and we can assume the electron did not bump into anything, or experience other, external forces along the way.

Each sphere should have uneven charge distribution because of induction. The field on the surface of sphere A closest to sphere B should be stronger than on the other side of sphere A, and the same applies to sphere B.

Yes, that's true! :)

Doesn't that affect the analysis?

Not in terms of the final answer. If the electron took a long path from the far end of one conductor to the far end of the other conductor [Edit: even if that means it takes a long, round-about path], the force acting on the electron would be less at any given position, but it would act over a longer displacement [integrated over the longer path], so the overall \int \vec F \cdot \vec {ds} would be the same as it would if it went straight from the near end of one conductor to the near end of the other (with the latter case being a greater force over a shorter displacement).

 

[collinsmark]

By the way, compare this problem to the old intro-physics type of problem where you have a block at some distance h above the ground. You can use conservation of energy to show that its energy when it reaches the ground is mgh. It doesn't matter if the block falls straight down, or if it slides down a shallow, frictionless incline, or even slides around on some weird, frictionless roller coaster sort of thing; it's energy is mgh when it reaches the ground regardless of the path it takes getting there (assuming there are no other forces such as friction).

[Edit: Not all forces/potentials have this property of being path independent. But gravity and electrostatic forces/potentials do. That's what makes them conservative.]

 

[alanf]

Thanks for your very clear answer (and sorry for being so late to reply). I think I have a better grasp of the geometry of the charge arrangement now.