Answer:
No
Step-by-step explanation:
The energy content of gasoline is
![44\cdot 10^6 J/kg](https://img.qammunity.org/2020/formulas/physics/high-school/rksm5td3agdeb7e7n7bk84aqilk7huljp5.png)
while the density is
![670 kg/m^3](https://img.qammunity.org/2020/formulas/physics/high-school/ivur19cr1luwxe7jmk4vjlyob2snwko5fe.png)
So the energy density of gasoline is
![u = (44\cdot 10^6 J/kg)(670 kg/m^3)=2.95\cdot 10^(10) J/m^3](https://img.qammunity.org/2020/formulas/physics/high-school/kitlltr3n2gmuhfp87xtcuecb9evwn3jv9.png)
The energy density of an electric field is given by
![u_E = (1)/(2)\epsilon_0 E^2](https://img.qammunity.org/2020/formulas/physics/high-school/cjt1co3xe0igwc4c3oukvr7v0ec2q6typp.png)
where
is the vacuum permittivity
E is the strength of the electric field
For air at dielectric breakdown,
![E=3 MV/m = 3\cdot 10^6 V/m](https://img.qammunity.org/2020/formulas/physics/high-school/u1s5u30771k43non9m4995x5xs9ycp9m9m.png)
Substituting into the equation,
![u_E = (1)/(2)(8.85\cdot 10^(-12))(3\cdot 10^6)^2=39.8 J/m^3](https://img.qammunity.org/2020/formulas/physics/high-school/8zxv4iwb7tdn520gwfd875k4ur16ws06np.png)
We see that
, so the energy density of the electric field is much lower than the energy content of gasoline, so the answer is No.