To find the dielectric constant \( k \) of the medium in the capacitor, we can follow these steps:
### Step 1: Write down the known values
- Capacitance \( C = 15 \, \text{pF} = 15 \times 10^{-12} \, \text{F} \)
- Voltage \( V = 500 \, \text{V} \)
- Electric Field \( E = 10^6 \, \text{V/m} \)
- Area of the plates \( A = 10^{-4} \, \text{m}^2 \)
- Permittivity of free space \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \)
### Step 2: Use the relationship between electric field, voltage, and distance
The electric field \( E \) is related to the voltage \( V \) and the distance \( d \) between the plates of the capacitor by the formula:
\[
E = \frac{V}{d}
\]
From this, we can express \( d \) as:
\[
d = \frac{V}{E}
\]
### Step 3: Substitute the known values to find \( d \)
Substituting the known values:
\[
d = \frac{500 \, \text{V}}{10^6 \, \text{V/m}} = 0.0005 \, \text{m} = 5 \times 10^{-4} \, \text{m}
\]
### Step 4: Use the capacitance formula
The capacitance \( C \) of a capacitor with a dielectric is given by:
\[
C = \frac{\epsilon A k}{d}
\]
Where:
- \( \epsilon = \epsilon_0 k \) (the permittivity of the dielectric material)
### Step 5: Rearrange the capacitance formula to solve for \( k \)
Rearranging the formula for \( k \):
\[
k = \frac{C d}{\epsilon_0 A}
\]
### Step 6: Substitute the known values into the formula for \( k \)
Substituting the values we have:
\[
k = \frac{(15 \times 10^{-12} \, \text{F})(5 \times 10^{-4} \, \text{m})}{(8.85 \times 10^{-12} \, \text{F/m})(10^{-4} \, \text{m}^2)}
\]
### Step 7: Calculate \( k \)
Calculating the numerator:
\[
15 \times 10^{-12} \times 5 \times 10^{-4} = 75 \times 10^{-16} \, \text{F m}
\]
Calculating the denominator:
\[
8.85 \times 10^{-12} \times 10^{-4} = 8.85 \times 10^{-16} \, \text{F}
\]
Now, substituting these into the equation for \( k \):
\[
k = \frac{75 \times 10^{-16}}{8.85 \times 10^{-16}} \approx 8.47
\]
### Conclusion
Thus, the dielectric constant \( k \) of the medium is approximately \( 8.47 \).
### Final Answer
The correct option is **(B) 8.47**.
