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Chapter 21: Problem 99

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.

Short Answer

Expert verified
The correct answer is (c) The plant became electrically polarized as the charged bee approached.

Step by step solution

01

Analyze Each Option

Consider each option carefully, understanding what it implies. Take into account the principles of physics concerning electrical polarization, the conductivity of air, and the behavior of charges.
02

Assess the Options

Option (a): Air is not a good conductor of electricity, hence it's not possible for positive charge from the bee's surface to flow through the air from bee to plant. Option (b): The earth is indeed a great reservoir of charges, but it mainly houses negative charges rather than positive ions. The movement of positive ions from earth to stem is unlikely. Option (d): This option is not factually correct, bees do not deposit positive charge on the plants.
03

Identify the Correct Option

Option (c): The plant can become electrically polarized as the charged bee approached. The positive charge on the bee will induce a redistribution of charges in the plant, leading to a positive charge on the part of the plant closest to the bee. This is the only option that is consistent with the principles of physics.

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Key Concepts

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

Electric Charge
An electric charge is a fundamental property of matter that occurs when there is an imbalance between protons and electrons in an atom. Protons carry a positive charge, while electrons carry a negative charge. When these charges are unequal, an object becomes charged. The concept of electric charge is central to understanding many phenomena in physics, including electric fields, forces, and polarization.
  • There are two types of electric charges: positive and negative.
  • Like charges repel each other, while opposite charges attract.
  • The movement or interaction of charges can result in noticeable forces.
In the context of the exercise, understanding how a positively charged bee approaching a plant leads to electricity's behavior is crucial. The charge on the bee induces a redistribution of charges in nearby objects, including the plant stem, which explains why the stem appears to become positively charged.
Conductivity of Air
Conductivity refers to a material's ability to allow the flow of electric charge. In simpler terms, it's how easily electricity can travel through a material. Air, however, is a poor conductor of electricity, meaning it does not easily allow electricity to pass through it. This is due to the large distance between air molecules, which makes the movement of charges less efficient.
  • Air has low conductivity because it's generally composed of non-conductive gases.
  • Only under extreme conditions (like lightning) can air become conductive.
  • In normal circumstances, electric charges cannot typically travel through air easily.
In the exercise, the suggestion that air could conduct charge from a bee to a plant is not feasible under ordinary conditions, reaffirming that air's poor conductivity does not support such charge transfer.
Charge Redistribution
Charge redistribution is a phenomenon where the presence of a charged object causes the charges in a neighboring object to rearrange. This process is a crucial aspect of electric polarization. When a charged object, such as our positively charged bee, nears another object like the plant, it can affect the charge distribution within the plant.
  • Redistribution occurs without the transfer of charges through a medium like air.
  • The electric field of the charged object influences the neighboring object's internal charges.
  • This can result in parts of the neighboring object becoming charged.
In the exercise, as the positively charged bee approaches the stem, it induces a charge separation in the stem. This means the charges in the plant rearrange, causing one side to become more positive, due to the influence of the electric field from the bee, explaining the observed positive charge.

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Most popular questions from this chapter

A very long line of charge with charge per unit length \(+8.00 \mu \mathrm{C} / \mathrm{m}\) is on the \(x\) -axis and its midpoint is at \(x=0 .\) A second very long line of charge with charge per length \(-4.00 \mu \mathrm{C} / \mathrm{m}\) is parallel to the \(x\) -axis at \(y=10.0 \mathrm{~cm}\) and its midpoint is also at \(x=0 .\) At what point on the \(y\) -axis is the resultant electric field of the two lines of charge equal to zero?

A point charge \(q_{1}=-4.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{~m}, y=0.800 \mathrm{~m},\) and a second point charge \(q_{2}=+6.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{~m}, y=0 .\) Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{C}\) is placed at the point \(x=+0.150 \mathrm{~m}, y=0,\) and an identical point charge is placed at \(x=-0.150 \mathrm{~m}, y=0 .\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{~m}, y=0\) (c) \(x=0.150 \mathrm{~m}, y=-0.400 \mathrm{~m}\) (d) \(x=0, y=0.200 \mathrm{~m}\)

Point charge \(q_{1}=-5.00 \mathrm{nC}\) is at the origin and point charge \(q_{2}=+3.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=3.00 \mathrm{~cm} .\) Point \(P\) is on the \(y\) -axis at \(y=4.00 \mathrm{~cm} .\) (a) Calculate the electric fields \(\vec{E}_{1}\) and \(\vec{E}_{2}\) at point \(P\) due to the charges \(q_{1}\) and \(q_{2}\). Express your results in terms of unit vectors (see Example 21.6 ). (b) Use the results of part (a) to obtain the resultant field at \(P\), expressed in unit vector form.

A negative charge of \(-0.550 \mu \mathrm{C}\) exerts an upward \(0.600 \mathrm{~N}\) force on an unknown charge that is located \(0.300 \mathrm{~m}\) directly below the first charge. What are (a) the value of the unknown charge (magnitude and sign); (b) the magnitude and direction of the force that the unknown charge exerts on the \(-0.550 \mu \mathrm{C}\) charge?
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