How much is the electric field inside your capacitor? E = _____V/m. How much is

Question

How much is the electric field inside your capacitor? E = _____V/m. How much is the total charge on your capacitor charged at its maximal voltage (30 kV)? Q = _____C. How much energy is stored inside of your capacitor? U = _____J. How much is the energy density in the electric field of your capacitor? U = _____J/m^3 = _____J/cm^3. Suppose, you want to use a capacitor (similar to the one you made) to store an energy for electric car. Gasoline has energy capacity about 120 MJ/gallon = _____J/cm^3 = _____J/m^3, so a typical 20-gallon gas tank can store about _____MJ = _____GJ of energy. About 30% of this energy, which is _____ MJ is converted by the internal combustion engine into the mechanical energy to drive the car (and the rest goes into heat which is mostly dissipated in the air passing via a radiator). Electric motors are much more efficient than gas engines (95-99%), so only 30% of gasoline energy is enough to drive electric car over the same distance as the gasoline car. How much is the volume of the capacitor capable of storing this amount (=30% of the 20-gallon gas tank) of energy? V = _____m^3 = _____gallons, which is ~_____times more than than typical 20 gallon tank. How big is this capacitor on a side (assuming it has a cube shape)? Side size = _____ m = _____ft. Current electrolytic capacitors have the energy density about ~ 10 J/cm^3. How bulky is such a capacitor for a typical electric car (how much is the volume of it)? _____m^3 = _____gallons, which is ~_____ times more than the typical 20 gallon gasoline car tank. The size of such capacitor (= the length of a side assuming a cubic shape) is still large:_____ m = _____ ft. Finally, some modem batteries have about ~1 KJ/cm^3 energy density, which would translate into the electric car battery of the volume _____ m^3 = _____ gallons and of the battery side size to be about _____ m = _____ cm = _____ ft (again assuming a cubic shape).

Explanation / Answer

Answering Question 2 ( with 5 parts)

a) 120 MJ/ gallon= 120 x 10^6/ 3785.41 J/ cm^3 = 31700.66 J /cm^3

b) 31700.66 J /cm^3 = 31700.66 J / 10^-6 m^3= 31700.66x 10^6 J/m^3

c) Energy stored by 20 gallon = 120 x 20 M J = 24 x 10^2 M J

d) 24 x 10^2 M J = 2.4 GJ

e) 30 % of MJ = 740 MJ

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