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 dielectric 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: the inner tube has a radius \( a \) and is maintained at a potential \( V \), while the outer tube has a radius \( b \) and is grounded. The space between the tubes is filled with dielectric oil of density \( \rho \) and dielectric constant \( K \).

### Step 2: Model the Capacitor
The system can be modeled as a capacitor with two regions: one filled with air (or vacuum) and the other filled with dielectric oil. The capacitance \( C \) of the system can be expressed as:
\[ C = C_{\text{air}} + C_{\text{oil}} \]
where \( C_{\text{air}} \) and \( C_{\text{oil}} \) are the capacitances of the air and oil regions, respectively.

### Step 3: Calculate Capacitances
The capacitance of the air region (between the inner tube and the outer tube) can be given by:
\[ C_{\text{air}} = \frac{2\pi \epsilon_0 L}{\ln(b/a)} \]
The capacitance of the oil region (which is filled with dielectric) can be expressed as:
\[ C_{\text{oil}} = \frac{2\pi K \epsilon_0 L}{\ln(b/a)} \]
Thus, the total capacitance becomes:
\[ C = \frac{2\pi \epsilon_0 L}{\ln(b/a)} + \frac{2\pi K \epsilon_0 L}{\ln(b/a)} = \frac{2\pi \epsilon_0 L (1 + K)}{\ln(b/a)} \]

### Step 4: Energy Stored in the Capacitor
The energy \( E \) stored in the capacitor can be expressed as:
\[ E = \frac{1}{2} C V^2 \]
Substituting for \( C \):
\[ E = \frac{1}{2} \cdot \frac{2\pi \epsilon_0 L (1 + K)}{\ln(b/a)} V^2 \]

### Step 5: Force on the Oil
The force acting on the oil column due to the electric field can be derived from the energy stored in the capacitor. The force \( F \) can be expressed as:
\[ F = \frac{dE}{dh} \]
We can relate this to the weight of the oil column:
\[ F = \rho g V_{\text{oil}} = \rho g \pi (b^2 - a^2) h \]
where \( V_{\text{oil}} \) is the volume of the oil column.

### Step 6: Equate Forces
At equilibrium, the upward force due to the electric field must balance the downward gravitational force on the oil:
\[ \frac{dE}{dh} = \rho g \pi (b^2 - a^2) h \]
Substituting for \( E \) and differentiating with respect to \( h \) gives us the equation to solve for \( h \).

### Step 7: Solve for \( h \)
After performing the necessary differentiation and rearranging, we can isolate \( h \) to find the equilibrium height:
\[ h = \frac{V^2 (1 + K)}{2 \rho g \ln(b/a)} \]

### Final Result
Thus, the equilibrium height \( h \) to which the oil rises in the space between the tubes is given by:
\[ h = \frac{V^2 (1 + K)}{2 \rho g \ln(b/a)} \]

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