Ignition: Potential Difference & Electric Field for Spark Plug

[sci0x]

Question: Spark plug in car has electrodes. They are spearated by 0.6mm. To create spark electric field of 3.0x10^6 V/m is needed.

a) What Potential difference is requuired to initiate the ignition.
Possible Answer: Is this just multiply Electric field by the distance between electrodes which would be -1800 J/C?

b) When separation between electrodes is increased, how does this affect the potential difference to initiate the spark? Explain
Possible Answer I'm guessing that potential difference will become more negative, thus decrease. I can't explain it though, could someone please.

c)Draw a graph that shows the magnitude of the electrical field as a function of the electrode separation.
Possible Answer Havn't done this yet, but I should get a straight line through the origin?

Thanks in advance

 

[Gamma]

Part a. ok
patt b. : VB - VA = - E.d where A is at a higher potantial than B. Hence,
(vB - vA) is a negative number. But we have added a negative sign in the right hand side to couter it. So when you increase d, the VA-VB will increase.

Plug in a value and see. say VB= 0V, VA= 50V, Then 50=Ed. If you increase d two fold. then (VB-VA)= -E 2d = -100 V

You can safely say that the magnitude of the potantial difference need to be increased to have a spark.

Partc:

not a straight line: Plot E vs. d. This is of the form xy = c^2. It's a rectangular hypabola.

 

[fatfatfat]

I have the same question with different numbers and I have a question about it.

deltaV = - E x d for a uniform electric field

My assignment accepted a positive answer. What happened to the negative sign?

Thanks :)