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Chapter 19: Problem 27

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of \(0.75 \mathrm{~mm}\). When an electric spark jumps between them, the magnitude of the electric field is \(4.7 \times 10^{7} \mathrm{~V} / \mathrm{m}\). What is the magnitude of the potential difference \(\Delta V\) between the conductors?

Short Answer

Expert verified
The magnitude of the potential difference is 35,250 V.

Step by step solution

01

Understand the Relationship

The electric field (E) is related to the potential difference (\Delta V) and the distance (d) between the conductors by the formula:\[ E = \frac{\Delta V}{d} \]
02

Rearrange the Formula

Rearrange the formula to solve for the potential difference:\[ \Delta V = E \times d \]
03

Plug in the Values

Substitute the given values of the electric field and the distance into the formula:\[ \Delta V = 4.7 \times 10^7 \, \mathrm{V/m} \times 0.75 \, \mathrm{mm} \]
04

Convert Millimeters to Meters

Since we need the units to match, convert the distance from millimeters to meters:\[ 0.75 \, \mathrm{mm} = 0.75 \times 10^{-3} \, \mathrm{m} \]
05

Calculate the Potential Difference

Now, calculate the magnitude of the potential difference:\[ \Delta V = 4.7 \times 10^7 \, \mathrm{V/m} \times 0.75 \times 10^{-3} \, \mathrm{m} = 3.525 \times 10^4 \, \mathrm{V} \]

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Key Concepts

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

Understanding the Electric Field
An electric field is a region around charged particles where other charges experience a force. Think of it as an invisible area where electrical forces take place. It's crucial in scenarios like a spark plug, where electric fields cause a spark to jump between conductors. The electric field is measured in volts per meter (V/m), indicating how much force a charge feels per meter in that field. Understanding this helps us grasp how a particular voltage leads to a spark in a given distance.
  • The formula linking electric field and potential difference is: \( E = \frac{\Delta V}{d} \).
  • This tells us the strength of the field depends on how much voltage is applied over a specific distance.
The Function of a Spark Plug
A spark plug is a device in engines responsible for igniting fuel. It creates a spark to start combustion by allowing electricity to jump a gap between two metal conductors. This spark ignites the air-fuel mixture, leading to the power needed to run an engine.
The spark plug鈥檚 gap needs a specific electric field strength to ignite鈥攁 crucial aspect of how vehicles run. It relies on the correct potential difference to make the spark jump, connecting this device closely to electric field principles.
  • Maintaining the correct gap ensures efficient ignition.
  • Regular checks and clean-ups help maintain performance.
Calculating Potential Difference
Potential difference, often called voltage, is the energy needed to move a charge between two points. In a spark plug, knowing this helps us understand the spark creation process. We've already seen the formula:
\[ \Delta V = E \times d \]
We input the known electric field and distance to find the potential difference. This basic calculation tells us how much energy is needed to push the current across the spark plug's gap. By using the given values, we find that:
\[ \Delta V = 4.7 \times 10^7 \, \mathrm{V/m} \times 0.75 \times 10^{-3} \, \mathrm{m} \]
Calculating gives us \(3.525 \times 10^4 \, \mathrm{V}\).
Understanding Unit Conversion
Unit conversion is changing a measurement to a different unit while keeping its value the same. It's crucial in physics to ensure calculations are correct鈥攍ike converting distances from millimeters to meters in our spark plug problem.
This conversion was necessary because the electric field was given in volts per meter, not volts per millimeter.
To convert millimeters to meters, remember:
  • 1 millimeter = 0.001 meters, or simply \(1 \, \mathrm{mm} = 10^{-3} \, \mathrm{m}\).
  • This allows consistent units across the calculation.

Without proper conversion, calculations could become erroneous, leading to incorrect conclusions about electrical systems.

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Most popular questions from this chapter

Concept Questions A positive point charge is surrounded by an equipotential surface \(A\), which has a radius of \(r_{A}\). A positive test charge moves from surface \(A\) to another equipotential surface \(B,\) which has a radius of \(r_{B} .\) In the process, the electric force does negative work. (a) Does the electric force acting on the test charge have the same or opposite direction as the displacement of the test charge? (b) Is \(r_{B}\) greater than or less than \(r_{A}\) ? Explain your answers. The positive point charge is \(q=+7.2 \times 10^{-8} \mathrm{C},\) and the test charge is \(q_{0}=+4.5 \times 10^{-11} \mathrm{C} .\) The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9} \mathrm{~J}\). The radius of surface \(A\) is \(r_{A}=1.8 \mathrm{~m} .\) Find \(r_{B}\) Check to see that your answer is consistent with your answers to the Concept Questions.

Two capacitors are identical, except that one is empty and the other is filled with a dielectric \((\kappa=4.50) .\) The empty capacitor is connected to a \(12.0-\mathrm{V}\) battery. What must be the potential difference across the plates of the capacitor filled with a dielectric such that it stores the same amount of electrical energy as the empty capacitor?

One particle has a mass of \(3.00 \times 10^{-3} \mathrm{~kg}\) and a charge of \(+8.00 \mu \mathrm{C}\). A second particle has a mass of \(6.00 \times 10^{-3} \mathrm{~kg}\) and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is \(0.100 \mathrm{~m},\) the speed of the \(3.00 \times 10^{-3}-\mathrm{kg}\) particle is \(125 \mathrm{~m} / \mathrm{s} .\) Find the initial separation between the particles.

Refer to Interactive Solution 19.39 at for one approach to this problem. The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of an \(850-\mu \mathrm{F}\) capacitor is \(280 \mathrm{~V}\). (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for \(3.9 \times 10^{-3} \mathrm{~s},\) find the effective power or "wattage" of the flash.

The electric potential energy stored in the capacitor of a defibrillator is \(73 \mathrm{~J}\), and the capacitance is \(120 \mu \mathrm{F}\). What is the potential difference that exists across the capacitor plates?
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