Potential Difference and spark plug

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SUMMARY

The discussion focuses on the calculations related to the potential difference required for a spark plug in a car, specifically with electrodes separated by a gap of 0.025 inches. To initiate a spark, an electric field of 3.0 x 106 V/m is necessary. The potential difference can be calculated using the formula delta V = -E delta s, where delta V is the potential difference, E is the electric field intensity, and delta s is the distance between the electrodes. Increasing the separation of the electrodes to 0.050 inches will require a higher potential difference to maintain the same electric field intensity.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Familiarity with the formula delta V = -E delta s
  • Knowledge of spark plug functionality in internal combustion engines
  • Basic concepts of charge and electric force (Coulomb's Law)
NEXT STEPS
  • Calculate the potential difference for a spark plug with a 0.050 inch gap
  • Explore the relationship between electric field intensity and electrode separation
  • Research the impact of varying electric fields on spark plug performance
  • Learn about the materials and design of spark plugs for optimal ignition
USEFUL FOR

Automotive engineers, electrical engineers, and students studying physics or automotive technology will benefit from this discussion, particularly those interested in ignition systems and spark plug design.

CFrazzle
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A spark plug in a car has electrodes separated by a gap of 0.025inches. To create a spark and ignite the air-fuel mixture in the engine, an electric field of 3.0 x 106 V/m is required in the gap. (a) What potential difference must be applied to the spark plug to initiate a spark? (b) If the separation between electrodes is increased, does the required potential difference increase, decrease, or stay the same. (c) Find the potential difference for a separation of 0.050”

delta V= -E delta s

E=Kq/r^2

r = distance
s = distance
delta V= potential difference
E=electric field intensity
K=9.0x10^9
q= charge


ive done the a section easy enough, but i have some questions about the b/c sections.
because the electrodes are seperating, will that also effect the electric field intensity.
 
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i have some questions about the b/c sections.
because the electrodes are seperating, will that also effect the electric field intensity.

Read the question: It is about the required potential difference - required to create a spark. "To create a spark, an electric field of 3.0 x 106 V/m is required in the gap." So what about the electric field intensity?

ehild
 

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