A parallel plate capacitor is charged by a DC supply of 500V. Plate separation is 1mm. If capacitoris lowered into water with water filling in the gap between the plates, then change in pressure at any point between the plates is `kepsilon_(0)xx10^(13)(N//m)^(2)`. Find the value of K, given `epsilon_r` (for water)=81.

Answer

To solve the problem, we need to analyze the change in pressure when a parallel plate capacitor is submerged in water. Here are the steps to find the value of \( K \):

### Step 1: Understand the relationship between pressure and energy density
The change in pressure (\( \Delta P \)) between the plates of the capacitor can be expressed in terms of energy density. The energy density (\( u \)) in an electric field is given by:
\[ u = \frac{1}{2} \epsilon E^2 \]
where \( \epsilon \) is the permittivity of the medium and \( E \) is the electric field.

### Step 2: Identify the initial and final conditions
Initially, the capacitor is in air, and when submerged in water, the permittivity changes. The final energy density in water can be expressed as:
\[ u_{\text{final}} = \frac{1}{2} \epsilon_r \epsilon_0 E^2 \]
where \( \epsilon_r \) is the relative permittivity of water (given as 81), and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Calculate the electric field
The electric field \( E \) between the plates of the capacitor can be calculated using the formula:
\[ E = \frac{V}{d} \]
where \( V = 500 \, \text{V} \) (the voltage) and \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \). Thus,
\[ E = \frac{500}{1 \times 10^{-3}} = 500000 \, \text{V/m} = 5 \times 10^5 \, \text{V/m} \]

### Step 4: Calculate the change in pressure
The change in pressure \( \Delta P \) when the capacitor is submerged in water can be expressed as:
\[ \Delta P = P_{\text{final}} - P_{\text{initial}} = \frac{1}{2} \epsilon_r \epsilon_0 E^2 - \frac{1}{2} \epsilon_0 E^2 \]
Factoring out \( \frac{1}{2} \epsilon_0 E^2 \):
\[ \Delta P = \frac{1}{2} \epsilon_0 E^2 (\epsilon_r - 1) \]

### Step 5: Substitute known values
Now we substitute \( \epsilon_r = 81 \) and \( E = 5 \times 10^5 \):
\[ \Delta P = \frac{1}{2} \epsilon_0 (5 \times 10^5)^2 (81 - 1) \]
Calculating \( (5 \times 10^5)^2 \):
\[ (5 \times 10^5)^2 = 25 \times 10^{10} = 2.5 \times 10^{11} \]
Substituting \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} \):
\[ \Delta P = \frac{1}{2} (8.854 \times 10^{-12}) (2.5 \times 10^{11}) (80) \]
Calculating:
\[ \Delta P = \frac{1}{2} (8.854 \times 10^{-12}) (2 \times 10^{13}) = 8.854 \times 10^{-12} \times 10^{13} = 8.854 \times 10^{1} = 88.54 \, \text{N/m}^2 \]

### Step 6: Relate to the given expression
The problem states that the change in pressure can be expressed as:
\[ \Delta P = k \epsilon_0 \times 10^{13} \, \text{N/m}^2 \]
Setting the two expressions for \( \Delta P \) equal gives:
\[ 88.54 = k \cdot 8.854 \times 10^{-12} \cdot 10^{13} \]
Solving for \( k \):
\[ k = \frac{88.54}{8.854 \times 10^{-12} \times 10^{13}} = \frac{88.54}{8.854} = 10 \]

### Final Answer
Thus, the value of \( K \) is:
\[ K = 10 \]