The discussion revolves around calculating the height (h) to which dielectric oil rises between two coaxial cylindrical metal tubes, with the inner tube at potential V and the outer grounded. Key equations include the capacitance formula for cylindrical capacitors, C/L = (2πkε₀)/(ln(b/a)), and the force equation F = (1/2)V²(dC/dx). Participants clarified the need to differentiate the capacitance with respect to height and equate forces for equilibrium, ultimately leading to the relationship between the electric force and the weight of the oil.
PREREQUISITESStudents and professionals in physics, electrical engineering, and materials science, particularly those dealing with electrostatics and dielectric materials.
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)}