Liquid Helium (Homework 7-323)

3. Suppose you heated a liter of helium liquid initially at 0.1°K by 4°K. Further suppose the charge-to-mass ratio e/m is unchanged during the heating (that is, e/m remains e/m0).
a. How would the total charge of the liquid change during heating?
b. If the liter of liquid were spherical, and all the excess charge migrated to the surface of the sphere, find the increase in electrostatic energy.
c. How would you modify the e/m law to ensure charge doesn’t change during heating?
Hints: Approximately, a liter of helium liquid becomes on vaporization 400 STP of gas. You can assume the kinetic energy is that of an ideal gas, and the gas atoms are moving non-relativistically.
This problem should help you appreciate that the total electric charge must be invariant.

Liquid Helium is usually not charged. However, suppose one has charged liquid Helium. Because of the electric filed generated by this charge (free) there will be also a polarization. Assume that total charge varies smoothly, i.e. continuous. At 4.1 K liquid Helium is just below its boiling point so that its charge in liquid state should be equal to its charge in gas state. In liquid state
$D=ϵ_0 E+P$  where $P=P=ϵ_0 χ*E$   ($χ+1=ϵ_r,ϵ_r=1.055$)
In gas state
$D=ϵ_0 E$     since He gas is a perfect gas ($ϵ_r=1.000065$ and $P≈0$)
In gas state charge is
$(Q(0.1)/V_{liquid} =)ρ=ρ_{free}=∇D$
In liquid state charge is
$(Q(4.1)/V_{gas} =)ρ=ρ_{free}+ρ_{induced}=∇D-∇P$    

Therefore (assume 1 L of liquid He)
$(Q(4.1)/200)/(Q(0.1)/1)=1-∇P/∇D=$
$=1-(ϵ_0 χ_{liquid} ∇E)/(ϵ_0 ∇E)=1-χ_{liquid}=1-0.055=0.945$

$Q(4.1)/Q(0.1) =0.945/200=0.0047$

The electrostatic energy is
$W=(QU^2)/2$
Given 1 L of helium the radius of the sphere is
$V=(4πR^3)/3$    and $R=∛(3V/4π)=∛((3*10^{-3})/4π)=0.062 m$
$U=Q/4πϵR$   and we can consider here that $ϵ(4.1)=ϵ(0.1)≈ϵ_0$

therefore $U(4.1)/U(0.1) =Q(4.1)/Q(0.1)$
$W(4.1)/W(0.1) =(Q^3 (4.1))/(Q^3 (0.1))=0.0047^3$    and
$ΔW/W(0.1) =(1-0.0047^3)/1≈1$
The energy stays the same.