Question

The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electric field of 3.0 x 10^6 V/m. What is the magnitude of the minimum potential difference that must be supplied by the ignition circuit to start a car

Answer 1

Answer:

The magnitude of minimum potential difference is 1800 V

Explanation:

Given:

Electric field $E = 3 \times 10^{6} \frac{V}{m}$

Gap between electrodes $d = 0.060 \times 10^{-2}$ m

For finding the minimum potential difference,

  $\Delta V = E \times d$

  $\Delta V = 3 \times 10^{6} \times 0.060 \times 10^{-2}$

  $\Delta V = 1800$ V

Therefore, the magnitude of minimum potential difference is 1800 V