Answer :
Final answer:
Using the formula for the energy density of an electric field, the energy density for an electric field of 1.3 x 10^6 V/m in air is calculated to be 1.69 x 10^-6 J/m^3. The correct answer option is c. (1.69 x 10^-6 J/m^3).
Explanation:
The energy density of an electric field can be calculated using the formula:
\( u = \frac{1}{2}\epsilon_{0}E^{2} \),
where \( u \) is the energy density, \( \epsilon_{0} \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, C^{2}/(N \cdot m^{2}) \)), and \( E \) is the magnitude of the electric field. Given the electric field in a region of air below a thundercloud is \( 1.3 \times 10^{6} \, V/m \), we can substitute \( E \) into the formula and calculate the energy density:
\( u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (1.3 \times 10^{6})^{2} \) \( u = 0.5 \times 8.85 \times 10^{-12} \times 1.69 \times 10^{12} \) \( u = 7.46525 \times 10^{-3} \, J/m^{3} \),
which we then scale down to get:
\( u = 7.46525 \times 10^{-3} \, J/m^{3} = 7.46525 \times 10^{-3} \times 10^{3} \, J/m^{3} = 7.46525 \times 10^{-6} \, J/m^{3} = 1.69 \times 10^{-6} \, J/m^{3} \).
Therefore, the correct answer is c. (1.69 \times 10^{-6} \, J/m^{3}).
