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Chapter 20: Problem 11

A spark plug in a car has electrodes separated by a gap of 0.025 in. To create a spark and ignite the air-fuel mixture in the engine, an electric field of \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m}\) 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? Explain. (c) Find the potential difference for a separation of 0.050 in.

Short Answer

Expert verified
a) 1905 volts. b) The potential difference increases. c) 3810 volts.

Step by step solution

01

Convert Measurements

First, convert the gap between electrodes from inches to meters. We know 1 inch equals 0.0254 meters, so:\[0.025 \text{ inches} = 0.025 \times 0.0254 \text{ meters} = 0.000635 \text{ meters}\]
02

Calculate the Potential Difference for 0.025 in Gap

Using the formula for electric field, \( E = \frac{V}{d} \), where \( V \) is the potential difference and \( d \) is the distance between the electrodes, rearrange to solve for \( V \):\[V = E \times d\]Given \( E = 3.0 \times 10^{6} \text{ V/m} \) and \( d = 0.000635 \text{ m} \), we find:\[V = 3.0 \times 10^{6} \text{ V/m} \times 0.000635 \text{ m} = 1905 \text{ volts}\]
03

Understand Effect of Increasing Electrode Separation

If the separation \( d \) increases while keeping the electric field \( E \) constant, according to the formula \( V = E \times d \), the required potential difference \( V \) increases. The reason is that a larger distance requires a greater voltage to maintain the same electric field strength.
04

Calculate the Potential Difference for 0.050 in Gap

Convert 0.050 inches to meters:\[0.050 \text{ inches} = 0.050 \times 0.0254 \text{ meters} = 0.00127 \text{ meters}\]Now, calculate the potential difference using \( E = 3.0 \times 10^{6} \text{ V/m} \) and \( d = 0.00127 \text{ m} \):\[V = 3.0 \times 10^{6} \text{ V/m} \times 0.00127 \text{ m} = 3810 \text{ volts}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Difference
In the context of electric fields, the potential difference is the voltage required to move an electric charge from one point to another within a field. In a car's spark plug, a potential difference is necessary to generate the electric spark needed to ignite the air-fuel mixture. You can think of the potential difference as the energy needed to push the electric charge across the gap between the electrodes.
The formula to look at is:
  • \( V = E \times d \)
where \( V \) is the potential difference, \( E \) is the electric field, and \( d \) is the distance between the electrodes. This equation shows how much voltage is necessary to maintain a specific electric field across a given distance.
By increasing the electric field strength or the distance, the potential difference increases. Hence, more voltage is required to maintain the electric field. This relationship is key in adjusting spark plug designs and ensuring proper engine function.
Electrode Separation
Electrode separation refers to the physical distance between the electrodes of a spark plug. This parameter is crucial because it directly affects the potential difference needed to create an electric spark. Imagine the electrode separation like stretching a rubber band 鈥� the more you stretch, the more effort it takes to cover the distance.
When we increase the separation:
  • The required potential difference increases, maintaining the same electric field.
  • More energy is needed to bridge the gap between the electrodes.
In our exercise, when the electrode gap increased from 0.025 inches to 0.050 inches, the potential difference needed almost doubled. Larger gaps in the spark plug mean that the ignition system has to work harder to create a spark. This is crucial for ensuring ignition and maintaining engine performance.
Voltage Calculation
Calculating the required potential difference (or voltage) for a spark plug involves understanding the formula that connects voltage, electric field, and electrode separation. The equation \( V = E \times d \) is our main tool here. It tells us how much voltage is necessary given a specific electric field and distance.
Steps for calculation:
  • Convert the gap from inches to meters if not already in metric units.
  • Multiply the electric field strength by the converted distance to find the voltage.
  • Adjust your understanding of results based on distance changes.
In the exercise, first, we found the gap in meters: 0.025 inches equals 0.000635 meters. Then, using the electric field strength 3.0 x 106 V/m, calculated the voltage as 1905 volts. Increasing the separation to 0.050 inches required recalculating, resulting in a higher voltage of 3810 volts. This calculation is essential for designing efficient and effective spark plugs.

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Most popular questions from this chapter

Lightning As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and a cloud at an altitude of \(550 \mathrm{m}\) to be the other plate. Assume the surface area of the cloud to be the same as the area of a square that is \(0.50 \mathrm{km}\) on a side. (a) What is the capacitance of this capacitor? (b) How much charge can the cloud hold before the dielectric strength of the air is exceeded and a spark (lightning) results?

In the Bohr model of the hydrogen atom, a proton and an electron are separated by a constant distance \(r .\) (a) Would the electric potential energy of the system increase, decrease, or stay the same if the electron is replaced with a proton? Explain. (b) Suppose, instead, that the proton is replaced with an electron. Would the electric potential energy of the system increase, decrease, or stay the same? Explain.

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a \(0.250-\mathrm{kg}\) ball with zero net charge was dropped from rest at a height of \(1.00 \mathrm{m}\). The ball landed 0.552 s later. Next, the ball was given a net charge of \(7.75 \mu \mathrm{C}\) and dropped in the same way from the same height. This time the ball fell for 0.680 s before landing. What is the electric potential at a height of \(1.00 \mathrm{m}\) above the ground on this planet, given that the electric potential at ground level is zero? (Air resistance can be ignored.)

Consider a region in space where a uniform electric field \(E=6500 \mathrm{N} / \mathrm{C}\) points in the negative \(x\) direction. (a) What is the orientation of the equipotential surfaces? Explain. (b) If you move in the positive \(x\) direction, does the electric potential increase or decrease? Explain. (c) What is the distance between the \(+14-V\) and the \(+16-V\) equipotentials?

Rutherford's Planetary Model of the Atom In 1911 , Ernest Rutherford developed a planetary model of the atom, in which a small positively charged nucleus is orbited by electrons. The model was motivated by an experiment carried out by Rutherford and his graduate students, Geiger and Marsden. In this experiment, they fired alpha particles with an initial speed of \(1.75 \times 10^{7} \mathrm{m} / \mathrm{s}\) at a thin sheet of gold. (Alpha particles are obtained from certain radioactive decays. They have a charge of \(+2 e\) and a mass of \(6.64 \times 10^{-27} \mathrm{kg}\).) How close can the alpha particles get to a gold nucleus (charge \(=+79 e\) ), assuming the nucleus remains stationary? (This calculation sets an upper limit on the size of the gold nucleus. See Chapter 31 for further details.)
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