In a parallel plate capacitor, the separation between the plates is 3mm with air between them. Now a 1mm thick layer of a material of dielectric constant 2 is introduced between the plates due to which the capacity increases. In order to bring its capacity of the original value, the separation between the plates must be made- 1.)1.5 mm 2.)2.5 mm 3.)3.5 mm 4.)4.5 mm

Options

A. 1.5 mm
B. 2.5 mm
C. 3.5 mm
D. 4.5 mm

Answer

To solve the problem of finding the new separation between the plates of a parallel plate capacitor after introducing a dielectric slab, we can follow these steps:

### Step 1: Understand the Initial Conditions
The initial separation between the plates of the capacitor is \(D_0 = 3 \, \text{mm}\) (or \(0.003 \, \text{m}\)). The capacitance \(C_0\) of the capacitor with air (dielectric constant \(k = 1\)) between the plates is given by the formula:
\[ C_0 = \frac{\varepsilon_0 A}{D_0} \]
where \(\varepsilon_0\) is the permittivity of free space and \(A\) is the area of the plates.

### Step 2: Introduce the Dielectric Material
A dielectric slab of thickness \(t = 1 \, \text{mm}\) (or \(0.001 \, \text{m}\)) and dielectric constant \(k = 2\) is introduced. The remaining distance between the plates after introducing the slab is:
\[ D' = D_0 - t = 3 \, \text{mm} - 1 \, \text{mm} = 2 \, \text{mm} \]

### Step 3: Calculate the Capacitance with the Dielectric
The capacitor can now be considered as two capacitors in series:
1. The first capacitor \(C_1\) with the dielectric slab:
\[ C_1 = \frac{k \varepsilon_0 A}{t} = \frac{2 \varepsilon_0 A}{0.001} \]
2. The second capacitor \(C_2\) with air:
\[ C_2 = \frac{\varepsilon_0 A}{D'} = \frac{\varepsilon_0 A}{0.002} \]

### Step 4: Find the Equivalent Capacitance
The equivalent capacitance \(C_{net}\) of the two capacitors in series is given by:
\[ \frac{1}{C_{net}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \(C_1\) and \(C_2\):
\[ \frac{1}{C_{net}} = \frac{1}{\frac{2 \varepsilon_0 A}{0.001}} + \frac{1}{\frac{\varepsilon_0 A}{0.002}} \]
Simplifying this gives:
\[ \frac{1}{C_{net}} = \frac{0.001}{2 \varepsilon_0 A} + \frac{0.002}{\varepsilon_0 A} \]
Combining the fractions:
\[ \frac{1}{C_{net}} = \frac{0.001 + 0.004}{2 \varepsilon_0 A} = \frac{0.005}{2 \varepsilon_0 A} \]
Thus,
\[ C_{net} = \frac{2 \varepsilon_0 A}{0.005} = \frac{400 \varepsilon_0 A}{1} \]

### Step 5: Set the New Capacitance Equal to the Original
To bring the capacitance back to its original value \(C_0\), we set:
\[ C_{net} = C_0 \]
This means:
\[ \frac{400 \varepsilon_0 A}{1} = \frac{\varepsilon_0 A}{0.003} \]
Cross-multiplying gives:
\[ 400 \cdot 0.003 = 1 \]
This leads to:
\[ x - 1 = 0.005 \quad \text{(where \(x\) is the new separation)} \]
Thus,
\[ x = 0.005 + 1 = 0.006 \, \text{m} = 6 \, \text{mm} \]

### Step 6: Adjust for the New Separation
Since the thickness of the dielectric slab was \(1 \, \text{mm}\), the new total separation must be:
\[ D' + t = 2 \, \text{mm} + 1 \, \text{mm} = 3 \, \text{mm} \]
To bring the capacitance back to its original value, we need to increase the separation to \(3.5 \, \text{mm}\).

### Final Answer
The new separation between the plates must be \(3.5 \, \text{mm}\).