91Ó°ÊÓ

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 ${\mathrm{χ}}_e$andmass density $\mathrm{ÒÏ}$.

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 $\mathrm{λ}$, inner radius $a$ and outer radius $b$ is

$\mathit{V}\mathbf{=}\frac{\mathbf{2}\mathbf{λ}}{\mathbf{4}\mathbf{Ï€}\mathbf{Î\mu }}\mathit{l}\mathit{n}\left(\frac{b}{a}\right)$ …… (1)

Here, $\mathrm{Î\mu }$is the permittivity of the medium.

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

Thus, from equation (1),

$\begin{array}{c}\frac{2\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\ln \left(\frac{b}{a}\right)=\frac{2{\mathrm{λ}}^{\text{'}}}{4\mathrm{Ï€}\mathrm{Î\mu }}\ln \left(\frac{b}{a}\right)\\ \frac{\mathrm{λ}}{{\mathrm{Î\mu }}_0}=\frac{{\mathrm{λ}}^{\text{'}}}{\mathrm{Î\mu }}\\ {\mathrm{λ}}^{\text{'}}={\mathrm{Î\mu }}_r\mathrm{λ}\end{array}$

Here, ${\mathrm{Î\mu }}_r$is the relative permittivity of the oil medium.

The net charge on the inner surface is

$Q={\mathrm{λ}}^{\text{'}}h+\mathrm{λ}(l−h)$

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

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

$\begin{array}{c}Q={\mathrm{Î\mu }}_r\mathrm{λ}h+\mathrm{λ}(l−h)\\ =\mathrm{λ}({\mathrm{χ}}_{e}h+l)\end{array}$

The expression for the capacitance in between the two surfaces is

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

Substitute the values in the above equation and get

$\begin{array}{c}C=\frac{\mathrm{λ}({\mathrm{χ}}_{e}h+l)}{\frac{2\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\ln \left(\frac{b}{a}\right)}\\ =2\mathrm{Ï€}{\mathrm{Î\mu }}_0\frac{({\mathrm{χ}}_{e}h+l)}{\ln \left(\frac{b}{a}\right)}\end{array}$

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

$\begin{array}{c}{F}_u=\frac{1}{2}{V}^2\frac{d}{dh}\left(2\mathrm{Ï€}{\mathrm{Î\mu }}_0\frac{({\mathrm{χ}}_{e}h+l)}{\ln \left(\frac{b}{a}\right)}\right)\\ =\frac{1}{2}{V}^2\frac{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\ln \left(\frac{b}{a}\right)}{\mathrm{χ}}_e\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\mathrm{.....}\left(2\right)\end{array}$

The expression for the downward gravitational force on the oil is

$F_d=\mathrm{ÒÏ}\mathrm{Ï€}gh(b^2−a^2)\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\mathrm{.....}\left(3\right)$

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 $\frac{{\mathrm{Î\mu }}_0{\mathrm{χ}}_e{V}^2}{\mathrm{ÒÏ}g(b^2−a^2)\ln \left(\frac{b}{a}\right)}$.