Question

A SPARK PLUG IN AN AUTOMOBILE ENGINE CONSISTS OF TWO

METALCONDUCTORS THAT ARE SPARATED BY A DISTANCE OF 0.75MM. WHEN
ANELECTRIC SPARK JUMPS BETWEEN THEM, THE MAGNITUDE OF THE
ELECTRICFIELD IS 4.7 X 10 TO THE 7TH V/M. WHAT IS THE MAGNITUDE OF
THEPOTENTIAL DIFFERENCE (DELTA V) BETWEEN THE CONDUCTORS?

Answer 1

Electric potential,
`V=3.52* 10^(12)\ volts`

Step-by-step explanation:

It is given that,

The magnitude of electric field,
`E=4.7* 10^7\ V/m`

Distance between automobile engines that consists of metal conductors,
`d = 0.75\ mm = 7.5* 10^4\ m`

Let V is the magnitude of the potential difference between the conductors. The relation between the electric field and the electric potential is given by:

`V=E* d`

`V=4.7* 10^7\ V/m * 7.5* 10^4\ m`

`V=3.52* 10^(12)\ volts`

So, the magnitude of the potential difference between the conductors is
`3.52* 10^(12)\ volts`. Hence, this is the required solution.