The separation between the plates of a parallel-plate capacitor of capacitance `20muF` is 2mm. The capacitor is initially uncharged. If a dielectric slab of surface area same as the capacitor, thickness `1mm`, and dielectric constant `2` is introduced, and then the new capacitance is?

Explanation

To find the new capacitance of the parallel-plate capacitor after introducing a dielectric slab, we can follow these steps:

### Step 1: Understand the Initial Setup
The initial capacitance \( C_0 \) of the capacitor is given as \( 20 \mu F \) with a plate separation \( D = 2 \text{ mm} \).

### Step 2: Calculate the Area of the Plates
Using the formula for capacitance:
\[ C = \frac{\varepsilon_0 A}{D} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( D \) is the separation between the plates. We can rearrange this to find the area \( A \):
\[ A = C_0 \cdot D / \varepsilon_0 \]
However, we do not need to calculate \( A \) explicitly since we will use it in the next steps.

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

### Step 4: Calculate the New Capacitance
The new capacitance \( C' \) can be calculated considering the two regions: one with the dielectric and one without. The capacitance of the section with the dielectric is given by:
\[ C_1 = \frac{k \varepsilon_0 A}{t} = \frac{2 \varepsilon_0 A}{1 \text{ mm}} \]
And the capacitance of the section without the dielectric is:
\[ C_2 = \frac{\varepsilon_0 A}{D'} = \frac{\varepsilon_0 A}{1 \text{ mm}} \]

Since these two capacitances are in series, the total capacitance \( C' \) can be found using the formula for capacitors in series:
\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} \]

Substituting the values:
\[ \frac{1}{C'} = \frac{1}{\frac{2 \varepsilon_0 A}{1 \text{ mm}}} + \frac{1}{\frac{\varepsilon_0 A}{1 \text{ mm}}} \]
\[ \frac{1}{C'} = \frac{1 \text{ mm}}{2 \varepsilon_0 A} + \frac{1 \text{ mm}}{\varepsilon_0 A} \]
\[ \frac{1}{C'} = \frac{1 + 2}{2 \varepsilon_0 A} = \frac{3}{2 \varepsilon_0 A} \]
Thus,
\[ C' = \frac{2 \varepsilon_0 A}{3} \]

### Step 5: Relate New Capacitance to Initial Capacitance
Since we know that \( C_0 = \frac{\varepsilon_0 A}{2 \text{ mm}} \), we can express \( \varepsilon_0 A \) in terms of \( C_0 \):
\[ \varepsilon_0 A = 2 \text{ mm} \cdot C_0 \]

Substituting this into the equation for \( C' \):
\[ C' = \frac{2 \cdot (2 \text{ mm} \cdot C_0)}{3} \]
\[ C' = \frac{4 \text{ mm}}{3} \cdot C_0 \]

### Step 6: Calculate \( C' \)
Now substituting \( C_0 = 20 \mu F \):
\[ C' = \frac{4 \text{ mm}}{3} \cdot 20 \mu F = \frac{80 \mu F}{3} \approx 26.67 \mu F \]

### Final Answer
The new capacitance \( C' \) is approximately \( 26.67 \mu F \).

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