The capacitance of a capacitor becomes` 4//3` times its original value if a dielectric slab of thickness `t=d//2` is inserted between the plates (d is the separation between the plates). What is the dielectric constant of the slab?

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we need to determine the dielectric constant \( k \) of a dielectric slab that is inserted between the plates of a capacitor, which increases the capacitance to \( \frac{4}{3} \) times its original value. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Original Capacitance
The original capacitance \( C_0 \) of a capacitor with a separation \( d \) between its plates and with vacuum (or air) as the dielectric is given by:
\[ C_0 = \frac{\varepsilon_0 A}{d} \]
where \( A \) is the area of the plates and \( \varepsilon_0 \) is the permittivity of free space.

### Step 2: Insert the Dielectric Slab
When a dielectric slab of thickness \( t = \frac{d}{2} \) is inserted, the remaining space between the plates is filled with air. Thus, the capacitor can be viewed as two capacitors in series:
1. The first capacitor (with the dielectric slab) has a capacitance \( C_1 \).
2. The second capacitor (with air) has a capacitance \( C_2 \).

### Step 3: Calculate the Capacitance with Dielectric
The capacitance \( C_1 \) of the portion with the dielectric slab is given by:
\[ C_1 = \frac{k \varepsilon_0 A}{\frac{d}{2}} = \frac{2k \varepsilon_0 A}{d} \]
where \( k \) is the dielectric constant of the slab.

### Step 4: Calculate the Capacitance with Air
The capacitance \( C_2 \) of the portion with air is:
\[ C_2 = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2 \varepsilon_0 A}{d} \]

### Step 5: Find the Equivalent Capacitance
The total capacitance \( C \) of two capacitors in series is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the expressions for \( C_1 \) and \( C_2 \):
\[ \frac{1}{C} = \frac{d}{2k \varepsilon_0 A} + \frac{d}{2 \varepsilon_0 A} \]
Combining the fractions:
\[ \frac{1}{C} = \frac{d(1 + k)}{2k \varepsilon_0 A} \]
Thus, the equivalent capacitance \( C \) becomes:
\[ C = \frac{2k \varepsilon_0 A}{d(1 + k)} \]

### Step 6: Set Up the Equation
According to the problem, the new capacitance \( C \) is \( \frac{4}{3} C_0 \):
\[ \frac{2k \varepsilon_0 A}{d(1 + k)} = \frac{4}{3} \cdot \frac{\varepsilon_0 A}{d} \]

### Step 7: Simplify the Equation
Canceling \( \varepsilon_0 A/d \) from both sides gives:
\[ 2k = \frac{4}{3}(1 + k) \]

### Step 8: Solve for \( k \)
Multiplying through by 3 to eliminate the fraction:
\[ 6k = 4(1 + k) \]
Expanding the right side:
\[ 6k = 4 + 4k \]
Rearranging gives:
\[ 6k - 4k = 4 \]
\[ 2k = 4 \quad \Rightarrow \quad k = 2 \]

### Final Answer
The dielectric constant \( k \) of the slab is:
\[ \boxed{2} \]