A parallel plate capacitor (condenser) has a certain capacitance (capacity). When 2/3 rd of the distance between the plates is filled with a dielectric, the (capacity) capacitance is found to be 2.25 times the initial capacitance. The dielectric constant of the dielectric

Options

A. 1
B. 3
C. 7
D. 6

Answer

To solve the problem of finding the dielectric constant \( k \) of a dielectric material that fills \( \frac{2}{3} \) of the distance between the plates of a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the initial capacitance
The initial capacitance \( C \) of a parallel plate capacitor with air (or vacuum) between the plates is given by:
\[ C = \frac{A \epsilon_0}{d} \]
where \( A \) is the area of the plates, \( \epsilon_0 \) is the permittivity of free space, and \( d \) is the distance between the plates.

### Step 2: Define the new configuration
When \( \frac{2}{3} \) of the distance \( d \) is filled with a dielectric of dielectric constant \( k \), the capacitor can be thought of as two capacitors in series:
1. Capacitor \( C_1 \) (with dielectric) has a distance of \( \frac{2}{3}d \).
2. Capacitor \( C_2 \) (without dielectric) has a distance of \( \frac{1}{3}d \).

### Step 3: Calculate the capacitances of the two sections
The capacitance of \( C_1 \) is:
\[ C_1 = \frac{A k \epsilon_0}{\frac{2}{3}d} = \frac{3A k \epsilon_0}{2d} \]
The capacitance of \( C_2 \) is:
\[ C_2 = \frac{A \epsilon_0}{\frac{1}{3}d} = 3A \epsilon_0 \]

### Step 4: Use the formula for capacitors in series
The total capacitance \( C' \) of the two capacitors in series is given by:
\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \( C_1 \) and \( C_2 \):
\[ \frac{1}{C'} = \frac{2d}{3A k \epsilon_0} + \frac{1}{3A \epsilon_0} \]

### Step 5: Find a common denominator
The common denominator is \( 3A k \epsilon_0 \):
\[ \frac{1}{C'} = \frac{2d + k d}{3A k \epsilon_0} = \frac{(2 + k)d}{3A k \epsilon_0} \]
Thus,
\[ C' = \frac{3A k \epsilon_0}{(2 + k)d} \]

### Step 6: Relate the new capacitance to the initial capacitance
According to the problem, the new capacitance \( C' \) is 2.25 times the initial capacitance \( C \):
\[ C' = 2.25 C \]
Substituting the expression for \( C' \):
\[ \frac{3A k \epsilon_0}{(2 + k)d} = 2.25 \cdot \frac{A \epsilon_0}{d} \]

### Step 7: Simplify the equation
Cancel \( A \epsilon_0/d \) from both sides:
\[ \frac{3k}{2 + k} = 2.25 \]

### Step 8: Solve for \( k \)
Cross-multiplying gives:
\[ 3k = 2.25(2 + k) \]
Expanding the right side:
\[ 3k = 4.5 + 2.25k \]
Rearranging:
\[ 3k - 2.25k = 4.5 \implies 0.75k = 4.5 \]
Thus,
\[ k = \frac{4.5}{0.75} = 6 \]

### Final Answer
The dielectric constant \( k \) is \( 6 \).

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