Capacitor, Dielectrics and height movement

[alphysicist]

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

Looks good to me. What do you get for the final value of h?

 

[TFM]

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:

$$h = \frac{(\frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) ))}{\rho*\left(\pi*b^2 - \pi*a^2\right)}$$

 

[alphysicist]

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:

$$h = \frac{(\frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) ))}{\rho*\left(\pi*b^2 - \pi*a^2\right)}$$

You are missing the factor of g. (You set the force equal to the mass, but it needs to be set equal to the weight.)