The plates of a parallel plate capacitor are charged to 100V. Then a 4mm thick dielectric slab is inserted between the plates and then to obtain the original potential difference , the distance between the system plates is increased by 2.00 mm. the dielectric constant of the slab is

Options

A. 5.4
B. 2.2
C. `2.0`
D. `2.6`

Answer

To solve the problem, we need to find the dielectric constant (k) of the dielectric slab inserted between the plates of a parallel plate capacitor. Here’s a step-by-step solution:

### Step 1: Understand the Initial Setup
The initial voltage (V) across the capacitor plates is given as 100V. The thickness of the dielectric slab (T) is 4 mm. When the dielectric slab is inserted, the distance between the plates is increased by 2 mm to maintain the original potential difference.

### Step 2: Define the Variables
- Let \( D \) be the original distance between the plates.
- After inserting the dielectric slab, the new distance \( D' \) becomes \( D + 2 \) mm.
- The thickness of the dielectric slab \( T = 4 \) mm.

### Step 3: Relate the Capacitance with and without Dielectric
The capacitance of a parallel plate capacitor without dielectric is given by:
\[ C = \frac{\epsilon_0 A}{D} \]
When the dielectric slab is inserted, the effective distance between the plates becomes \( D' = D + 2 \) mm, and the capacitance with the dielectric is:
\[ C' = \frac{\epsilon_0 A}{D - T} \cdot k \]
Where \( k \) is the dielectric constant of the slab.

### Step 4: Set Up the Equation
Since the potential difference remains the same (100V), we can equate the capacitances:
\[ \frac{Q}{V} = C = \frac{\epsilon_0 A}{D} \quad \text{and} \quad \frac{Q}{V} = C' = \frac{\epsilon_0 A k}{D + 2 - T} \]
This gives us:
\[ \frac{\epsilon_0 A}{D} = \frac{\epsilon_0 A k}{D + 2 - T} \]

### Step 5: Cancel Out Common Terms
We can cancel \( \epsilon_0 A \) from both sides (assuming they are non-zero):
\[ \frac{1}{D} = \frac{k}{D + 2 - T} \]

### Step 6: Rearranging the Equation
Cross-multiplying gives:
\[ D + 2 - T = kD \]
Rearranging this, we get:
\[ kD = D + 2 - T \]
Substituting \( T = 4 \) mm:
\[ kD = D + 2 - 4 \]
Which simplifies to:
\[ kD = D - 2 \]

### Step 7: Solve for k
Now, we can express k as:
\[ k = \frac{D - 2}{D} \]

### Step 8: Determine the Value of D
Since the problem does not provide the original distance \( D \), we can assume a value for \( D \). However, we know that \( D \) must be greater than 4 mm (the thickness of the dielectric slab). For simplicity, let’s assume \( D = 6 \) mm:
\[ k = \frac{6 - 2}{6} = \frac{4}{6} = \frac{2}{3} \]

### Step 9: Conclusion
Thus, the dielectric constant \( k \) of the slab is \( \frac{2}{3} \).