The plates of a parallel plate capacitor are charged up to `100 v`. Now, after removing the battery, a `2 mm` thick plate is inserted between the plates Then, to maintain the same potential deffernce, the distance betweem the capacitor plates is increase by `1.6 mm`. The dielectric canstant of the plate is .

Options

A. 6
B. 8
C. 5
D. 4

Answer

To find the dielectric constant \( K \) of the plate inserted between the capacitor plates, we can follow these steps:

### Step 1: Understand the Initial Conditions
The initial potential difference \( V \) across the capacitor is given as \( 100 \, \text{V} \). The distance between the plates is \( D \).

### Step 2: Capacitance of the Original Capacitor
The capacitance \( C_0 \) of the original capacitor can be expressed as:
\[ C_0 = \frac{\varepsilon_0 A}{D} \]
where \( \varepsilon_0 \) is the permittivity of free space and \( A \) is the area of the plates.

### Step 3: Charge on the Capacitor
Since the battery is removed, the charge \( Q \) on the capacitor remains constant:
\[ Q = C_0 V = \frac{\varepsilon_0 A}{D} \cdot 100 \]

### Step 4: Inserting the Dielectric Plate
A dielectric plate of thickness \( t = 2 \, \text{mm} \) is inserted between the plates. The distance between the plates is then increased by \( 1.6 \, \text{mm} \). Thus, the new distance between the plates becomes:
\[ D' = D + 1.6 \, \text{mm} \]

### Step 5: New Capacitance with Dielectric
The new configuration consists of two capacitors in series:
1. The capacitor with the dielectric \( C_1 \):
\[ C_1 = \frac{K \varepsilon_0 A}{t} \]
2. The capacitor without the dielectric \( C_2 \):
\[ C_2 = \frac{\varepsilon_0 A}{D' - t} = \frac{\varepsilon_0 A}{(D + 1.6) - 2} \]

### Step 6: Total Capacitance in Series
The total capacitance \( C \) of the system is given by the formula for capacitors in series:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]

Substituting the expressions for \( C_1 \) and \( C_2 \):
\[ \frac{1}{C} = \frac{t}{K \varepsilon_0 A} + \frac{(D + 1.6 - 2)}{\varepsilon_0 A} \]
Simplifying gives:
\[ \frac{1}{C} = \frac{t}{K \varepsilon_0 A} + \frac{D - 0.4}{\varepsilon_0 A} \]

### Step 7: Equating Capacitances
Since the potential difference remains the same, we have:
\[ C = C_0 \]
Thus:
\[ \frac{1}{C_0} = \frac{1}{C_1} + \frac{1}{C_2} \]

### Step 8: Substitute and Solve for K
Substituting \( C_0 = \frac{\varepsilon_0 A}{D} \):
\[ \frac{D}{\varepsilon_0 A} = \frac{K}{t} + \frac{D - 0.4}{\varepsilon_0 A} \]
Cross multiplying and simplifying leads to:
\[ D = K \cdot \frac{D - 0.4}{D} \]
From this equation, we can isolate \( K \) and solve:
\[ 0.4K = 2 \implies K = 5 \]

### Final Answer
The dielectric constant \( K \) of the plate is \( 5 \).

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