Question

Two long coaxial tubes (with inner radius $a$ and outer rdius $b$) are placed vertically into a container with oil (relative permittivity ${\epsilon }_r$, density $\rho$). The inner tube has potential, $U$ the second is grounded. What is the height $h$ in which the oil between tubes ascends to?
Use this equation : cylindrical capacitor with length $l$ will inner radius $a$ and outer radius $b$ filled $b$ filled by dielectric with relative permittivity ${\epsilon }_r$ has a capacity $C$
$C=\frac{2\pi \epsilon l}{ln\left(b/a\right)}$

Answer 1

Given: Two long coaxial tubes with radius a and radius b are placed vertically
into a container with oil.
According to the formula,
cylindrical capacitor with length $l$ inner radius $a$
and outer radius $b$ filled by dielectric with relative permittivity
$ϵ\left(r\right)$.
$C=\frac{(2\times \pi \times ϵ0\times l)}{\ln \left(\frac{b}{a}\right)}$
Total value in this case of a capacitor can be given as:
$C=C\left(air\right)+C\left(oil\right)$
$=\frac{(2\times \pi \times ϵ0\times x)}{\ln \left(\frac{b}{a}\right)}+\frac{(2\times \pi \times ϵr\times (l-x))}{\ln \left(\frac{b}{a}\right)}$
$=\frac{(2\times \pi \times ϵ0\times [(ϵr-1)\times x+1)\left]\right)}{\ln \left(\frac{b}{a}\right)}$
This equation follows the fact that the two capacitors are placed in parallel.
When the surface of the oil increases by value $dr$
capacity of the capacitor changes to:
$dc=\frac{(2\times \pi \times ϵ0\times [(ϵr-1)\times (x+dx)+1)\left]\right)}{\ln \left(\frac{b}{a}\right)}-\frac{(2\times \pi \times ϵ0\times [(ϵr-1)\times x+1)\left]\right)}{\ln \left(\frac{b}{a}\right)}$
$dc=\frac{(2\times \pi \times ϵ0\times [(ϵr-1)\times dx])}{\ln \left(\frac{b}{a}\right)}$
The energy stored in the capacitor
$dE=\frac{1}{2}\times U^2\times dc$
Energy stored is work done
Hence,
$dE=dW=F\times dx$
So we have calculated the force acting upon:
$F=\frac{1}{2}\times U^2\times \left(\frac{dc}{dx}\right)$
= $\frac{1}{2}\times U^2\times \frac{(2\times \pi \times ϵ0\times (ϵr-1))}{\ln \frac{b}{a}}$
We calculated the effect of electric forces on dielectric. These forces draw the dielectric inside the capacitor up to such a height which is limited by gravity force.
$F=m\times g$
= $e\times \pi \times (b^2-a^2)\times g\times h$
From which, $h=\frac{\left(ϵ0\right)\times (ϵr-1)\times U^2}{e\times (b^2-a^2)\times g\times \ln \frac{b}{a}}$