IP A spark plug in a car has electrodes separated by a gap of 6.5×10−2 in . To create a spark and ignite the air-fuel mixture in the engine, an electric field of 3.2×106 V/m is required in the gap.

a.) What potential difference must be applied to the spark plug to initiatea spark?b.) Would the necessary applied voltage be greater or less if theseparation were greater? Would that be a good idea for animprovement in design? Find the potential difference for a separationof 0.05 inches.

There are several conditions that must be met for a spark to be created into an air gap. Once the physical conditions are fixed, a minimum electric field is necessary for the spark to be initiated. Let s be the separation between the electrodes and V their potential difference. The electric field is

a.)

[tex]\displaystyle E=\frac{V}{s}[/tex]

Solving for V

[tex]V=E.s[/tex]

The separation is

[tex]s= 5.5\cdot 10^{-2}\ in=0.001651 \ m[/tex]

Thus the potential difference is

[tex]V=3.2\cdot 10^{6}\ V/m\times 0.001651 \ m[/tex]

[tex]V=5283.2 \ V[/tex]

b.) If the separation was greater, the applied voltage needs to be greater if the electric field has to be constant. One possible measure to keep electrodes as close as possible is to build them as sharp edges. It gives the spark an easier path to travel to.

If the separation is 0.05 inches =0.00127 m

[tex]V=3.2\cdot 10^{6}\ V/m\times 0.00127 \ m[/tex]

[tex]V=4064\ V[/tex]

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-02-19T06:23:07+01:00

(a) a potential difference of 5283 V must be applied to the spark plug to initiate a spark

(b) It will be lesser

Potential Difference

(a) Let r be the separation between the electrodes and V their potential difference. The electric field E is given, so

V = E.r

here r = 6.5×10⁻² in = 6.5×10⁻²×0.0254 m

and E = 3.2×10⁶ V/m

[tex]V=3.2\times10^6\times6.5\times10^{-2}\times0.0254V\\\\V=5283V[/tex]

(b) If the separation is 0.05 inches =0.00127 m, then the potential difference required is:

[tex]V = 3.2\times10^6\times0.00127\;V\\\\V=4064V[/tex]

which is less than the voltage required in the original condition.

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