Two long coaxial cylindrical metal tubes (inner radius a,outer radiusb)stand vertically in a tank of dielectric oil (susceptibility ${\mathbf{\chi }}_e$,mass density $\mathbf{\rho }$).The inner one is maintained at potential V,and the outer one is grounded (Fig. 4.32). To what height (h) does the oil rise, in the space between the tubes?

There are two long coaxial cylindrical metal tubes of inner radius aandouter radius

b.

The tubesstand vertically in a tank of dielectric oil having susceptibility ${\chi }_e$andmass density $\rho$.

The inner cylinder is maintained at potential Vand the outer one is grounded.

The potential in between a coaxial cylindrical space with line charge density $\lambda$, inner radius $a$ and outer radius $b$ is

$\mathit{V}\mathbf{=}\frac{\mathbf{2}\mathbf{\lambda }}{\mathbf{4}\mathbf{\pi }\mathbf{\epsilon }}\mathit{l}\mathit{n}\left(\frac{b}{a}\right)$ …… (1)

Here, $\epsilon$is the permittivity of the medium.

Let $\lambda$and $\lambda \text{'}$be the charge densities on the inner surface corresponding to the air and oil medium and ${\epsilon }_0$ and $\epsilon$ be their permittivity's. The potential difference between the two surfaces remains constant.

Thus, from equation (1),

$\begin{array}{c}\frac{2\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{b}{a}\right)=\frac{2{\lambda }^{\text{'}}}{4\pi \epsilon }\ln \left(\frac{b}{a}\right)\\ \frac{\lambda }{{\epsilon }_0}=\frac{{\lambda }^{\text{'}}}{\epsilon }\\ {\lambda }^{\text{'}}={\epsilon }_r\lambda \end{array}$

Here, ${\epsilon }_r$is the relative permittivity of the oil medium.

The net charge on the inner surface is

$Q={\lambda }^{\text{'}}h+\lambda (l-h)$

Here, $l$ is the total height of the cylinder.

Substitute the expression for $\lambda \text{'}$in the above equation

$\begin{array}{c}Q={\epsilon }_r\lambda h+\lambda (l-h)\\ =\lambda ({\chi }_{e}h+l)\end{array}$

The expression for the capacitance in between the two surfaces is

$C=\frac{Q}{V}$

Substitute the values in the above equation and get

$\begin{array}{c}C=\frac{\lambda ({\chi }_{e}h+l)}{\frac{2\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{b}{a}\right)}\\ =2\pi {\epsilon }_0\frac{({\chi }_{e}h+l)}{\ln \left(\frac{b}{a}\right)}\end{array}$

The expression for the net upward force is

$F_u=\frac{1}{2}{V}^2\frac{dC}{dh}$

Substitute the values in the above equation and get

$\begin{array}{c}{F}_u=\frac{1}{2}{V}^2\frac{d}{dh}\left(2\pi {\epsilon }_0\frac{({\chi }_{e}h+l)}{\ln \left(\frac{b}{a}\right)}\right)\\ =\frac{1}{2}{V}^2\frac{2\pi {\epsilon }_0}{\ln \left(\frac{b}{a}\right)}{\chi }_e\mathrm{.....}\left(2\right)\end{array}$

The expression for the downward gravitational force on the oil is

$F_d=\rho \pi gh(b^2-a^2)\mathrm{.....}\left(3\right)$

At equilibrium the net upward force should be equal to the net downward force.

Thus, equate equations (2) and (3)

Thus, the height till which the oil rises is $\frac{{\epsilon }_0{\chi }_e{V}^2}{\rho g(b^2-a^2)\ln \left(\frac{b}{a}\right)}$.