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]

Options

A. `(epsilon_(0)2V^(2)(K-1))/(grho(b^(2)-a^(2))1n((b)/(a)))`
B. `(epsilon_(0)2V^(2)(K-1))/(rho(b^(2)-a^(2))g1n((b)/(a)))`
C. `(4epsilon_(0)2V^(2)(K-1))/(grho(b^(2)-a^(2))1n((b)/(a)))`
D. `(6epsilon_(0)2V^(2)(K-1))/(rho(b^(2)-a^(2))g1n((b)/(a)))`

Answer

To solve the problem of determining the equilibrium height (h) to which the oil rises in the space between two coaxial cylindrical tubes, we can follow these steps:

### Step 1: Understand the System
We have two coaxial cylindrical tubes: an inner tube with radius \( a \) and an outer tube with radius \( b \). The inner tube is maintained at a potential \( V \), while the outer tube is grounded (potential = 0). The space between the tubes is filled with dielectric oil of mass density \( \rho \) and dielectric constant \( K \).

### Step 2: Determine the Capacitance
The capacitance \( C \) of a cylindrical capacitor filled with a dielectric can be expressed as:
\[ C = \frac{2 \pi \epsilon_0 \epsilon_r L}{\ln(b/a)} \]
where \( \epsilon_r = K \) (the dielectric constant of the oil), and \( L \) is the length of the cylinders.

### Step 3: Calculate the Electric Field
The electric field \( E \) in the region between the two cylinders can be calculated using the relationship between potential difference and electric field:
\[ E = \frac{V}{d} \]
where \( d \) is the distance between the two cylinders, which is \( b - a \).

### Step 4: Force on the Dielectric
The force acting on the dielectric oil due to the electric field can be expressed as:
\[ F = \frac{1}{2} \epsilon E^2 A \]
where \( A \) is the cross-sectional area of the inner cylinder, given by \( A = \pi a^2 \).

### Step 5: Weight of the Oil
The weight of the oil column of height \( h \) is given by:
\[ W = \rho g V_o \]
where \( V_o \) is the volume of oil, \( V_o = A \cdot h = \pi a^2 h \).

### Step 6: Equilibrium Condition
At equilibrium, the upward electric force must balance the weight of the oil:
\[ \frac{1}{2} \epsilon E^2 A = \rho g \cdot \pi a^2 h \]

### Step 7: Substitute and Rearrange
Substituting \( E \) and \( A \) into the equilibrium condition:
\[ \frac{1}{2} \cdot \epsilon_0 K \cdot \left(\frac{V}{\ln(b/a)}\right)^2 \cdot \pi a^2 = \rho g \cdot \pi a^2 h \]
We can cancel \( \pi a^2 \) from both sides:
\[ \frac{1}{2} \cdot \epsilon_0 K \cdot \left(\frac{V}{\ln(b/a)}\right)^2 = \rho g h \]

### Step 8: Solve for Height \( h \)
Rearranging gives us:
\[ h = \frac{\epsilon_0 K \cdot V^2}{2 \rho g \cdot \ln(b/a)} \]

### Final Answer
Thus, the equilibrium height \( h \) to which the oil rises is:
\[ h = \frac{\epsilon_0 K \cdot V^2}{2 \rho g \cdot \ln(b/a)} \]