Voltage rating of a parallel plate capacitor is `500V`. Its dielectric can withstand a maximum electric field of `10^(6) V//m`. The plate area is `10^(-4)m^(2)`. What is the dielectric constant if the capacitance is `15pF`?
(given `epsilon_(0)=8.86xx10^(-12)C^(2)//Nm^(2)`)

Options

A. `3.8`
B. `8.5`
C. `4.5`
D. `6.2`

Answer

To find the dielectric constant \( K \) of the parallel plate capacitor, we can follow these steps:

### Step 1: Determine the separation \( D \) between the plates
The electric field \( E \) in a capacitor is given by the formula:
\[ E = \frac{V}{D} \]
where \( V \) is the voltage rating of the capacitor. Rearranging this formula gives:
\[ D = \frac{V}{E} \]
Given:
- \( V = 500 \, \text{V} \)
- \( E = 10^6 \, \text{V/m} \)

Substituting the values:
\[ D = \frac{500 \, \text{V}}{10^6 \, \text{V/m}} = 500 \times 10^{-6} \, \text{m} = 5 \times 10^{-4} \, \text{m} \]

### Step 2: Use the capacitance formula to find \( K \)
The capacitance \( C \) of a parallel plate capacitor filled with a dielectric is given by:
\[ C = \frac{K \epsilon_0 A}{D} \]
Rearranging this formula to solve for \( K \) gives:
\[ K = \frac{C D}{\epsilon_0 A} \]

### Step 3: Substitute the known values into the formula
Given:
- \( C = 15 \, \text{pF} = 15 \times 10^{-12} \, \text{F} \)
- \( \epsilon_0 = 8.86 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( A = 10^{-4} \, \text{m}^2 \)
- \( D = 5 \times 10^{-4} \, \text{m} \)

Now substituting these values into the equation for \( K \):
\[ K = \frac{(15 \times 10^{-12} \, \text{F}) \cdot (5 \times 10^{-4} \, \text{m})}{(8.86 \times 10^{-12} \, \text{C}^2/\text{N m}^2) \cdot (10^{-4} \, \text{m}^2)} \]

### Step 4: Calculate \( K \)
Calculating the numerator:
\[ 15 \times 10^{-12} \cdot 5 \times 10^{-4} = 75 \times 10^{-16} \, \text{F m} \]

Calculating the denominator:
\[ 8.86 \times 10^{-12} \cdot 10^{-4} = 8.86 \times 10^{-16} \, \text{C}^2/\text{N} \]

Now substituting these into the equation for \( K \):
\[ K = \frac{75 \times 10^{-16}}{8.86 \times 10^{-16}} \approx 8.47 \]

Thus, the dielectric constant \( K \) is approximately \( 8.5 \).

### Final Answer:
The dielectric constant \( K \) is approximately \( 8.5 \).

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