The plates of a parallel plate capacitor are charged up to 200 V. A dielectric slab of thickness 4 mm is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates increased by 3.2 mm. The dielectric constant of the dielectric slab is

Options

A. 1
B. 4
C. 5
D. 6

Explanation

To solve the problem step by step, we will analyze the situation involving a parallel plate capacitor with a dielectric slab inserted between its plates.

### Step 1: Understand the initial conditions
We have a parallel plate capacitor charged to a potential difference \( V = 200 \, V \). The initial distance between the plates is \( d \).

### Step 2: Insert the dielectric slab
A dielectric slab of thickness \( t = 4 \, mm \) is inserted between the plates. The dielectric constant of the slab is denoted as \( k \).

### Step 3: Change in distance between the plates
After inserting the dielectric slab, the distance between the plates is increased by \( 3.2 \, mm \). Therefore, the new distance between the plates becomes:
\[ d' = d + 3.2 \, mm \]

### Step 4: Relate the capacitance before and after inserting the dielectric
The capacitance of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{d} \]
Where:
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.

When the dielectric is inserted, the capacitance becomes:
\[ C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{k}} \]
Where \( \frac{t}{k} \) accounts for the effective reduction in distance due to the dielectric.

### Step 5: Charge remains constant
When the dielectric is inserted, the charge \( Q \) on the capacitor remains constant. Thus, we can equate the two expressions for charge:
\[ Q = C \cdot V = C' \cdot V' \]
Since \( V' \) remains 200 V, we can write:
\[ \frac{\varepsilon_0 A}{d} \cdot 200 = \frac{\varepsilon_0 A}{d' - t + \frac{t}{k}} \cdot 200 \]
This simplifies to:
\[ \frac{1}{d} = \frac{1}{d' - t + \frac{t}{k}} \]

### Step 6: Substitute the values
From the problem, we know:
- \( t = 4 \, mm \)
- \( d' = d + 3.2 \, mm \)

Substituting these into the equation gives:
\[ \frac{1}{d} = \frac{1}{(d + 3.2) - 4 + \frac{4}{k}} \]

### Step 7: Rearranging the equation
This leads to:
\[ d - (d + 3.2 - 4 + \frac{4}{k}) = 0 \]
Which simplifies to:
\[ 3.2 = 4 - \frac{4}{k} \]

### Step 8: Solve for \( k \)
Rearranging gives:
\[ \frac{4}{k} = 4 - 3.2 = 0.8 \]
Thus:
\[ k = \frac{4}{0.8} = 5 \]

### Final Answer
The dielectric constant \( k \) of the dielectric slab is \( 5 \).

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