Question

The gap between electrodes in a spark plug is d, 0.05 cm (not meters). To produce an electric spark in a gasoline-air mixture, there must be an electric field, E, of 3.5 x 10^6 V/m. What minimal potential difference, V, must be supplied by the ignition to start a car

Answer 1

Answer:

The voltage that must be supplied is

V=1750volts

Explanation:

Step one :

The formula for the electric field between the charges is given by

E=V/d

Where E= the electric field

V= voltage or potential difference

d= the distance between the two electrodes

Step two :

Given that

E=3.5*0^6v/m

d=0.05cm to meter we have 0.0005m

V= unknown

By making v subject of formula we can solve for the required voltage

V=d*E

V=0.0005*3.5*10^6

V=1750volts

Answer 2

Answer:

The minimal potential difference that must be supplied by the ignition to start the car is 1750 V

Explanation:

Given data:

E = 3.5x10⁶V/m

d = 0.05 cm = 5x10⁻⁴m

The minimun voltage is equal:

$V_{min} =Ed=3.5x10^{6} *5x10^{-4} =1750V$