Two square metal plates of side `1 m` are kept `0.01 m` apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicualr to an oil surface in the a tank filled with an insulating oil. The plates are connected to a battery of emf `500 V`. The plates are then lowered vertically into the oil at a speed of `0.001 ms^(-1)`. Calculate the current drawn from the battery during the process.
(Dielectric constant of oil `=11, epsilon_(0) = 8.85 xx 10^(-12) N^(-1) m^(-2))`.

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Configuration
We have two square metal plates of side 1 m, separated by a distance of 0.01 m, acting as a parallel plate capacitor. The plates are connected to a battery of emf 500 V and are lowered into an insulating oil at a speed of 0.001 m/s.

### Step 2: Identify Given Values
- Side of the plates, \( L = 1 \, \text{m} \)
- Separation between the plates, \( D = 0.01 \, \text{m} \)
- Dielectric constant of oil, \( K = 11 \)
- Emf of the battery, \( V = 500 \, \text{V} \)
- Speed of lowering plates, \( \frac{dx}{dt} = 0.001 \, \text{m/s} \)
- Permittivity of free space, \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{N}^{-1} \text{m}^{-2} \)

### Step 3: Determine Capacitance
As the plates are lowered, part of them will be in air and part in oil. Let \( x \) be the length of the plates submerged in oil, then the length in air is \( L - x \).

The capacitance of the part in air (denoted as \( C_1 \)) and the part in oil (denoted as \( C_2 \)) can be calculated as follows:

1. **Capacitance in Air:**
\[ C_1 = \frac{\epsilon_0 \cdot L \cdot (L - x)}{D} \]

2. **Capacitance in Oil:**
\[ C_2 = \frac{K \cdot \epsilon_0 \cdot L \cdot x}{D} \]

### Step 4: Total Capacitance
The total capacitance \( C \) is the sum of \( C_1 \) and \( C_2 \):
\[ C = C_1 + C_2 = \frac{\epsilon_0 \cdot L}{D} \left( (L - x) + K \cdot x \right) \]
\[ C = \frac{\epsilon_0 \cdot L}{D} \left( L + (K - 1)x \right) \]

### Step 5: Calculate Charge and Current
The charge \( Q \) on the capacitor is given by:
\[ Q = C \cdot V \]
The current \( I \) drawn from the battery can be expressed as:
\[ I = \frac{dQ}{dt} = \frac{d(C \cdot V)}{dt} \]
Using the product rule:
\[ I = V \frac{dC}{dt} + C \frac{dV}{dt} \]
Since \( V \) is constant, \( \frac{dV}{dt} = 0 \), so:
\[ I = V \frac{dC}{dt} \]

### Step 6: Differentiate Capacitance
To find \( \frac{dC}{dt} \), we note that \( x \) changes with time:
\[ \frac{dC}{dt} = \frac{\epsilon_0 \cdot L}{D} (K - 1) \frac{dx}{dt} \]

### Step 7: Substitute Values
Now we substitute the known values into the equation for current:
\[ I = V \cdot \frac{\epsilon_0 \cdot L}{D} (K - 1) \frac{dx}{dt} \]
Substituting the values:
\[ I = 500 \cdot \frac{8.85 \times 10^{-12} \cdot 1}{0.01} (11 - 1) \cdot 0.001 \]
\[ I = 500 \cdot \frac{8.85 \times 10^{-12}}{0.01} \cdot 10 \cdot 0.001 \]
\[ I = 500 \cdot 8.85 \times 10^{-12} \cdot 1 \]
\[ I = 4.425 \times 10^{-9} \, \text{A} \]

### Final Answer
The current drawn from the battery during the process is approximately:
\[ I \approx 4.425 \, \text{nA} \]