Question

Two long coaxial cylindrical metal tubes (inner radius $a$ , outer radius $b$ ) stand vertically in a tank of dielectric oil (of mass density $\rho $ , dielectric constant $K$ ). The inner one is maintained at potential $V$ and the outer one is grounded. To what equilibrium height ( $h$ ) does the oil rise in the space between the tubes? [Assume this height ( $h$ ) as an equilibrium height]

• A. $\frac{\varepsilon_0 2 V^2(K-1)}{g \rho\left(b^2-a^2\right) \ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)}$
• B. $\frac{\varepsilon_0 V^2(K-1)}{\rho\left(b^2-a^2\right) g \ln \left(\frac{b}{a}\right)}$
• C. $\frac{4 \varepsilon_0 V^2(K-1)}{g \rho\left(b^2-a^2\right) \ln \left(\frac{b}{a}\right)}$
• D. $\frac{6 \varepsilon_0 \mathrm{~V}^2(\mathrm{~K}-1)}{\rho\left(\mathrm{b}^2-\mathrm{a}^2\right) \mathrm{g} \ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)}$

Answer 1

Answer: (B)

$Gravitational \, force=mg$
$= \, \rho \pi \left(b^{2} - \, a^{2}\right)gh$
Net upward force $F \, =\frac{d U}{d h}$
$=\frac{1}{2}V^{2} \, \frac{d C}{d h}$
where $h$ is the height of liquid.
Now we calculate $C$ as a function of $h$




$C_{e q} \, = \, C_{A i r} \, + \, C_{K}$
$=\frac{2 \pi \varepsilon_0(L-h)}{\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)}+\frac{2 \pi \varepsilon_0 \mathrm{Kh}}{\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)}$
$C=\frac{2 \pi \varepsilon_0\{L+(K-1) h\}}{\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)}$
$\mathrm{F}=\frac{1}{2} \mathrm{~V}^2 \times \frac{\mathrm{dC}}{\mathrm{dh}}=\frac{1}{2} \mathrm{~V}^2 \times \frac{2 \pi \varepsilon_0(\mathrm{~K}-1)}{\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)}$
$\therefore \mathrm{F}=\mathrm{mg}$
$\frac{\frac{1}{2} \mathrm{~V}^2 \times 2 \pi \varepsilon_0(\mathrm{~K}-1)}{\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)}=\rho \pi\left(\mathrm{b}^2-\mathrm{a}^2\right) \mathrm{gh}$
$h=\frac{\pi \varepsilon_0 V^2(K-1)}{\rho \pi\left(b^2-a^2\right) g \ln \left(\frac{b}{a}\right)}$
$=\frac{\varepsilon_0 \mathrm{~V}^2(\mathrm{~K}-1)}{\rho\left(\mathrm{b}^2-\mathrm{a}^2\right) \mathrm{g} \ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)}$