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Which falls faster in a plasma, a conductor or an insulator?

+ 2 like - 0 dislike
521 views

Suppose you stand immersed in a large, earthbound chamber filled with a net-neutral proton-electron plasma. You hold two spheres that are identical in every way except for their conductivities. The first sphere is a perfect conductor and the second sphere is a perfect insulator. If you drop the spheres, which one falls faster?

asked May 30, 2015 in Theoretical Physics by Josh Burby (120 points) [ no revision ]

1 Answer

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My tentative answer is that the conducting sphere falls faster, but there may be another subtle effect that I'm missing.

The important difference between the two spheres (conducting and insulating), is that the conducting sphere must maintain a constant potential across its surface, which essentially requires the sheath to be the same width at the top and bottom. In contrast, the movement of the insulating sphere means the sheath is larger on the top than the bottom, essentially creating a polarization drag as it falls. 

In a little more detail, let's work out the size of the sheath for the insulating sphere. Neither sphere draws any net current so we can calculate this by equating the ion and electron currents.  The ion current is simply given by the Bohm criterion -- ions move into the object at the Bohm velocity -- giving \[j_i=en_e (c_s\pm v_s).\]Here \(c_s =\sqrt{k_B T_e/m_i}\) is the ion sound speed (Bohm velocity), \(e\) is the electron charge, \(n_e\) and \(T_e\) are the electron density and temperature, \(m_i\) is the ion mass and \(v_s\) is the speed of the sphere. The + accounts for the extra ion flux on the bottom of the sphere, while the - accounts for decreased flux on the top (as it moves downwards).  

The electron current is given by assuming a Maxwellian distribution function and integrating up to the potential of the sphere, \(\Delta V\). This gives \[j_e = \sqrt{\frac{k_B T_e}{2\pi m_e}}e^{-\frac{e\Delta V}{k_B T_e}}.\]We then equate ion and electron currents to solve for \(\Delta V\), giving\[\Delta V=\frac{k_B T_e}{2e} \ln \left( \frac{m_i}{2\pi m_e} \left(1\mp \frac{2v_s}{c_s} \right)\right), \]where the - is for the bottom and the + is for the top. Thus, we see that there is a larger potential difference between the plasma and the top of the sphere than the plasma and the bottom of the sphere. The electric field associated with this is pointing towards the sphere, which is negatively charged, meaning there is a larger upwards force on the top than downwards force on the bottom -- i.e., a drag. 

For the conducting sphere, charges will move around on the surface so as to exactly keep the potential constant, counteracting this effect. Thus it falls as if in a neutral gas. 

answered May 31, 2015 by JonoSquire (40 points) [ revision history ]
edited Jun 3, 2015 by JonoSquire

If we switch off gravity, does that mean that a moving in plasma object can be decelerated by plasma due to this "drag" force?

@VladimirKalitvianski Yes, the effect arises purely from the difference in velocity between the plasma ions and the object, not so differently to a normal drag force on any object moving in a gas. 

@JonoSquire As far as I understand, if an insulator is put in plasma, first it is the fast electrons that are deposited on the insulator surface and charge it negatively. Then they attract ions and repel the plasma electrons, so some flux of ions is established towards the surface. The plasma neutrality is broken within the Debye length. Thus the insulator "consumes" plasma staying slightly charged negatively. The local Debye length depends on the local plasma density, so "behind" the body it may be slightly different from the front length and the electric force of plasma-body interaction may be different in principle. The resulting electric force is added to the usual drag force arising in case of a moving object in some gas.

In case of a perfectly conducting body, the situation is similar except for the electric field which is uniform along the surface, so no net electric force arises and the usual drag force is the only force acting on a moving conductor. This is what you are saying. But equal electric field may act on different (local) electric charge on the conducting surface. So I have some doubts about your conclusions.

@VladimirKalitvianski I think that in the case of the conductor, the surface charge on the sphere will move around so that the electron charge density is everywhere equal. This is the only way for the voltage difference between the plasma and the sphere to be the same everywhere on the surface (since fundamentally the charge density is responsible for the voltage difference). In the case of the insulator there will be more surface charge density on the top of the sphere, where the electric field is also greater, so there is no possibility that the two effects (surface charge density and normal electric field magnitude) will cancel each other out. Note that a current flows through the conducting sphere as it falls, to account for the unequal electron and ion currents on either side of the sphere.

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