/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 During a lightning flash, a pote... [FREE SOLUTION] | 91Ó°ÊÓ

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During a lightning flash, a potential difference \(V\) cloud \(-V\) ground exists between a cloud and the ground. As a result, negative electric charge is transferred from the ground to the cloud. (a) As the charge moves from the ground to the cloud, work is done on the charge by the electric force. How is this work related to the potential difference and the charge \(q ?(\mathrm{~b})\) If this work could be used to accelerate an automobile of mass \(m\) from rest, what would be the automobile's final speed? Express your answer in terms of the potential difference, the charge, and the mass of the automobile. (c) If this work could all be converted into heat, what mass \(m\) of water could be heated so that its temperature increases by \(\Delta T ?\) Write your answer in terms of the potential difference, the charge, the mass, the specific heat capacity \(c\) of water, and \(\Delta T\) Suppose a potential difference of \(V_{\text {cloud }}-V_{\text {ground }}=1.2 \times 10^{9} \mathrm{~V}\) exists between the cloud and the ground, and \(q=-25 \mathrm{C}\) of charge is transferred from the ground to the cloud. (a) How much work \(W\) ground-cloud is done on the charge by the electric force? (b) If the work done by the electric force were used to accelerate a 1100 \(\mathrm{kg}\) automobile from rest, what would be its final speed? (c) If the work done by the electric force were converted into heat, how many kilograms of water at \(0{ }^{\circ} \mathrm{C}\) could be heated to \(100^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
(a) \(-3.0 \times 10^{10} \text{ J}\). (b) \(7.35 \times 10^3 \text{ m/s}\). (c) \(7172 \text{ kg}\).

Step by step solution

01

Calculate Work Done by Electric Force

The work done by the electric force on a charge moving through a potential difference is given by the formula \( W = q \times V \). Here, \( q = -25 \text{ C} \) and \( V = 1.2 \times 10^9 \text{ V} \). Thus, the work done is \( W = (-25) \times (1.2 \times 10^9) = -3.0 \times 10^{10} \text{ J} \). The negative sign indicates work done in moving a negative charge.
02

Determine Final Speed of the Automobile

To find the final speed of an automobile accelerated by this work, use the work-energy principle. The equation \( W = \frac{1}{2}mv^2 \) relates work to kinetic energy. Rearrange for \( v \): \( v = \sqrt{\frac{2W}{m}} \). Substituting \( W = 3.0 \times 10^{10} \text{ J} \) and \( m = 1100 \text{ kg} \), we get \( v = \sqrt{\frac{2 \times 3.0 \times 10^{10}}{1100}} \approx 7.35 \times 10^3 \text{ m/s} \).
03

Calculate Mass of Water Heated

The heat \( Q \) produced if the work is converted into heat is equal to the work done: \( Q = W \). To find the mass \( m \) of water heated, use the equation \( Q = mc\Delta T \). Here, \( c = 4186 \text{ J/kg°C} \) (the specific heat capacity of water), and \( \Delta T = 100 \text{ °C} \). Solving \( m = \frac{Q}{c\Delta T} = \frac{3.0 \times 10^{10}}{4186 \times 100} \approx 7172 \text{ kg} \).

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Key Concepts

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

Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. When an electric charge moves through a potential difference, work is done by the electric force. This work can be calculated using the formula \( W = q \times V \), where \( q \) is the charge and \( V \) is the potential difference. Since the work represents energy, it can be used to change the state of a system. For instance, if this work were applied to a stationary automobile, it would translate into the car's kinetic energy, causing it to move with a certain speed. Thus, understanding the work-energy principle helps connect the dots between energy transformations in electric systems and mechanical work.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It is given by the equation \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In the context of the lightning flash exercise, if the work done by the electric force were transferred to an automobile, this work would appear as the car's kinetic energy.
  • The total work done translates directly into the car's kinetic energy.
  • To find the car's speed, rearrange the kinetic energy formula to solve for \( v \).
If an automobile starts from rest and is subjected to a force that does a certain amount of work, its speed can be calculated using these formulas, showing the practical applications of kinetic energy.
Specific Heat Capacity
Specific heat capacity is a property that indicates how much heat energy is required to change the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). The formula for calculating the heat transferred to a substance is \( Q = mc\Delta T \), where \( Q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.
  • The specific heat capacity of water is high, known to be about \( 4186 \text{ J/kg°C} \).
  • This high value means it takes a lot of energy to change the water's temperature, making it an effective coolant.
In this exercise, if the work done by the electric force is entirely converted to heat, one could assess how much water can be warmed from 0°C to 100°C using the specific heat capacity of water.
Electric Force
Electric force is the force exerted by electrically charged objects on each other. It is responsible for the work done when moving charges through a potential difference. This fundamental force can be attractive or repulsive depending on the nature of the charges involved.
  • An electric field associated with a potential difference \( V \) influences this force.
  • The amount of work done by the electric force on charge \( q \) is derived as \( W = q \times V \).
In practical terms, electric force not only influences charged particles but is also harnessed to perform work in various applications ranging from electronics to large systems like automobiles, illustrating its versatile role in energy transfer.
Potential Difference
Potential difference, commonly referred to as voltage, is the work needed to move a charge from one point to another in an electric field. It is measured in volts (V) and is a key factor in determining how much work an electric force can do.
  • The potential difference exists due to an electric field created, for example, between a cloud and the ground.
  • It drives the charge movement, which is accompanied by the electric force doing work.
A potential difference essentially represents the energy transfer capability of an electric field, playing a crucial role in determining how charged particles behave and establishing power across circuits or natural phenomena like lightning.

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Most popular questions from this chapter

Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is \(0 \mathrm{~V} ?\)

An electron is released at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing), (a) As the electron gains kinetic energy, does its electric potential energy increase or decrease? Why? (b) The difference in the electron's electric potential energy between the positive and negative plates is EPE positive \(-\mathrm{EPE}_{\text {negative. }}\) How is this difference related to the charge on the electron \((-e)\) and to the difference \(V\) positive \(-V_{\text {negative in the electric potential between }}\) the plates? (c) How is the potential difference \(V\) positive \(-V_{\text {negative }}\) related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plate? The plates of a parallel plate capacitor are separated by a distance of \(1.2 \mathrm{~cm}\), and the electric field within the capacitor has a magnitude of \(2.1 \times 10^{6} \mathrm{~V} / \mathrm{m}\). An electron starts from rest at the negative plate and accelerates to the positive plate. What is the kinetic energy of the electron just as the electron reaches the positive plate?

Two particles each have a mass of \(6.0 \times 10^{-3} \mathrm{~kg}\). One has a charge of \(+5.0 \times 10^{-6} \mathrm{C}\), and the other has a charge of \(-5.0 \times 10^{-6} \mathrm{C}\). They are initially held at rest at a distance of \(0.80 \mathrm{~m}\) apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-third its initial value?

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.

What voltage is required to store \(7.2 \times 10^{-5} \mathrm{C}\) of charge on the plates of a \(6.0-\mu \mathrm{F}\) capacitor?

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