Answer:
No
Explanation:
The energy content of gasoline is
[tex]44\cdot 10^6 J/kg[/tex]
while the density is
[tex]670 kg/m^3[/tex]
So the energy density of gasoline is
[tex]u = (44\cdot 10^6 J/kg)(670 kg/m^3)=2.95\cdot 10^{10} J/m^3[/tex]
The energy density of an electric field is given by
[tex]u_E = \frac{1}{2}\epsilon_0 E^2[/tex]
where
[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity
E is the strength of the electric field
For air at dielectric breakdown,
[tex]E=3 MV/m = 3\cdot 10^6 V/m[/tex]
Substituting into the equation,
[tex]u_E = \frac{1}{2}(8.85\cdot 10^{-12})(3\cdot 10^6)^2=39.8 J/m^3[/tex]
We see that [tex]u_E < u[/tex], so the energy density of the electric field is much lower than the energy content of gasoline, so the answer is No.
