To find the dielectric constant (k) of the capacitor, we can follow these steps:
### Step 1: Calculate the Distance Between the Plates
The electric field (E) in a parallel plate capacitor is given by the formula:
\[
E = \frac{V}{d}
\]
where \( V \) is the voltage across the plates and \( d \) is the distance between the plates.
Given:
- Voltage rating \( V = 500 \, V \)
- Maximum electric field \( E = 10^6 \, V/m \)
Rearranging the formula to find \( d \):
\[
d = \frac{V}{E}
\]
Substituting the values:
\[
d = \frac{500 \, V}{10^6 \, V/m} = 5 \times 10^{-4} \, m
\]
### Step 2: Use the Capacitance Formula
The capacitance \( C \) of a parallel plate capacitor with a dielectric is given by:
\[
C = k \cdot \frac{\epsilon_0 \cdot A}{d}
\]
where:
- \( k \) is the dielectric constant,
- \( \epsilon_0 = 8.86 \times 10^{-12} \, C^2/(N \cdot m^2) \) (permittivity of free space),
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.
Given:
- Capacitance \( C = 15 \, pF = 15 \times 10^{-12} \, F \)
- Area \( A = 10^{-4} \, m^2 \)
### Step 3: Rearranging for Dielectric Constant (k)
Rearranging the capacitance formula to solve for \( k \):
\[
k = \frac{C \cdot d}{\epsilon_0 \cdot A}
\]
### Step 4: Substitute the Values
Substituting the known values into the equation:
\[
k = \frac{(15 \times 10^{-12} \, F) \cdot (5 \times 10^{-4} \, m)}{(8.86 \times 10^{-12} \, C^2/(N \cdot m^2)) \cdot (10^{-4} \, m^2)}
\]
### Step 5: Calculate k
Calculating the numerator:
\[
15 \times 10^{-12} \cdot 5 \times 10^{-4} = 75 \times 10^{-16} \, F \cdot m
\]
Calculating the denominator:
\[
8.86 \times 10^{-12} \cdot 10^{-4} = 8.86 \times 10^{-16} \, C^2/N \cdot m
\]
Now substituting back:
\[
k = \frac{75 \times 10^{-16}}{8.86 \times 10^{-16}} \approx 8.47
\]
### Final Answer
The dielectric constant \( k \) is approximately \( 8.47 \).
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