The problem involves two long coaxial cylindrical metal tubes placed vertically in a tank of dielectric oil. The inner tube is maintained at a potential V while the outer tube is grounded. The question seeks to determine the height to which the oil rises between the tubes.
The discussion is ongoing, with participants actively exploring different aspects of the problem. Some have provided guidance on how to approach the derivation of capacitance and the necessary forces, while others are still clarifying their understanding of the relationships between the variables involved.
Participants note that the problem involves multiple interpretations of the setup, particularly regarding the lengths of the capacitor sections in and out of the oil. There is also mention of needing to express certain lengths in terms of the height of the oil, indicating a need for careful consideration of the geometry involved.
TFM said:I though it seemed to small...
So:
(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})*(1+\chi_e)
(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})*(1+\chi_e)
Goes to:
\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})} + \chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})
so
F = \frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) )
?
TFM
TFM said:F also = mg, so
\rho * \left(\pi*b^2*h - \pi*a^2*h\right) = \frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) )
take the h out the brackets:
\rho *h \left(\pi*b^2 - \pi*a^2\right) = \frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) )
giving:
h = \frac{(\frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) ))}{\rho*\left(\pi*b^2 - \pi*a^2\right)}