/*! 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 101 In a follow-up experiment, a cha... [FREE SOLUTION] | 91Ó°ÊÓ

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In a follow-up experiment, a charge of \(+40\) pC was placed at the center of an artificial flower at the end of a 30-cm long stem. Bees were observed to approach no closer than 15 cm from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? (a) \(2.4 \space N/C; (b) 16 \space N/C; (c) 2.7 \times 10^{-10} \space N/C; (d) 4.8 \times 10^{-10} \space N/C.\)

Short Answer

Expert verified
The smallest external electric field is 16 N/C.

Step by step solution

01

Understand the Problem

We need to determine the smallest external electric field that deters the bee from approaching closer than 15 cm to a point charge of +40 pC. The options are given in N/C.
02

Identify Key Equations

We can calculate the electric field at a certain distance from a point charge using Coulomb's law: \[ E = \frac{k \cdot |q|}{r^2} \] where \( E \) is the electric field, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \), \( |q| \) is the magnitude of the charge, and \( r \) is the distance from the charge.
03

Insert Known Values

Here, \( |q| = 40 \, \text{pC} = 40 \times 10^{-12} \, \text{C} \) and \( r = 15 \, \text{cm} = 0.15 \, \text{m} \). Substitute these into the equation: \[ E = \frac{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 40 \times 10^{-12} \, \text{C}}{(0.15 \, \text{m})^2} \]
04

Perform the Calculation

Calculate the electric field: \[ E = \frac{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 40 \times 10^{-12} \, \text{C}}{0.0225 \, \text{m}^2} \] \[ E = \frac{359.6 \times 10^{-3} \, \text{N m}^2/\text{C}}{0.0225 \, \text{m}^2} \] \[ E \approx 15.98 \, \text{N/C} \]
05

Compare with Given Options

The calculated electric field \( 15.98 \, \text{N/C} \) is closest to option (b) 16 N/C.

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Key Concepts

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

Coulomb's Law
Coulomb's law is a fundamental principle in physics that describes the electrostatic interaction between two point charges. It states that the electric force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This law is represented mathematically as:
  • Force (\[ F \]) between two charges: \[ F = \frac{k \cdot |q_1 q_2|}{r^2} \]
where
  • \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
  • \( |q_1| \) and \( |q_2| \) are the magnitudes of the charges
  • \( r \) is the distance between the centers of the two charges
This law helps in calculating the strength and direction of the electrostatic force, which is essential in understanding electric fields and the behavior of charged particles. By understanding how charges interact, we can predict the force acting on any charge placed within an electric field created by other charges.
Point Charge
A point charge is an idealized model of a charged particle in which all the charge is concentrated at a single point in space. This simplification allows for straightforward calculations of electric fields and forces, as it reduces complex charge distributions to a simpler, ideal scenario. In physics problems like the one described, point charges serve as a useful approximation when the dimensions of the charged body are much smaller than the distances involved in the problem. This concept is crucial for applying Coulomb's law and determining the electric fields produced by such charges.When dealing with point charges, the electric field created by a point charge can be easily calculated at any given distance using the formula:
  • Electric field (\[ E \]) due to a point charge: \[ E = \frac{k \cdot |q|}{r^2} \]
Here:
  • \( |q| \) is the magnitude of the point charge
  • \( r \) is the distance from the charge to the point where the field is being calculated
A point charge acts as the source of an electric field, which extends outward in all directions. This concept is vital in the problem of determining the distance at which bees sense the electric field from the artificial flower.
Electric Field Calculation
The electric field is a vector field that represents the force exerted by a charged object on a test charge placed within the field. To determine the strength of an electric field produced by a point charge, we use the equation derived from Coulomb's law:
  • \[ E = \frac{k \cdot |q|}{r^2} \]
This equation enables us to calculate the magnitude of the electric field at any given distance \( r \) from the charge \( q \).For the exercise given:
  • Charge \( |q| = 40 \, \text{pC} = 40 \times 10^{-12} \, \text{C} \)
  • Distance \( r = 15 \, \text{cm} = 0.15 \, \text{m} \)
Substituting these values into the electric field formula, you perform the calculation as follows:First, substitute known values into the equation:\[ E = \frac{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 40 \times 10^{-12} \, \text{C}}{0.15^2 \, \text{m}^2} \]Simplify the expression to find:\[ E = \frac{359.6 \times 10^{-3} \, \text{N m}^2/\text{C}}{0.0225 \, \text{m}^2} \]Resulting in:\[ E \approx 15.98 \, \text{N/C} \]This calculated electric field strength helps us understand the threshold beyond which bees react to the presence of an external electric field.

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Most popular questions from this chapter

A small sphere with mass \(m\) carries a positive charge \(q\) and is attached to one end of a silk fiber of length \(L\). The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density \(\sigma\). Show that when the sphere is in equilibrium, the fiber makes an angle equal to arctan (\(q\sigma/2mg\epsilon_0\)) with the vertical sheet.

The earth has a net electric charge that causes a field at points near its surface equal to 150 N\(/\)C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth's electric field a feasible means of flight? Why or why not?

A nerve signal is transmitted through a neuron when an excess of \(Na^+\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately 10.0 \(\mu\)m in diameter, and measurements show that about 5.6 \(\times \space 10^{11} \space Na^+\)ions per meter (each of charge \(+e\)) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of point charges moving along the axon. Consider a 0.10-mm length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.10-mm length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 cm below the skin? (c) Certain sharks can respond to electric fields as weak as 1.0 \(\mu N/C\). How far from this segment of axon could a shark be and still detect its electric field?

Two very large parallel sheets are 5.00 cm apart. Sheet \(A\) carries a uniform surface charge density of \(-8.80 \space \mu\)C\(/m^2\), and sheet \(B\), which is to the right of \(A\), carries a uniform charge density of \(-11.6 \space \mu\)C\(/m^2\). Assume that the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet \(A\); (b) 4.00 cm to the left of sheet \(A\); (c) 4.00 cm to the right of sheet \(B\).

A ring-shaped conductor with radius \(a =\) 2.50 cm has a total positive charge \(Q = +\)0.125 nC uniformly distributed around it (see Fig. 21.23). The center of the ring is at the origin of coordinates \(O\). (a) What is the electric field (magnitude and direction) at point \(P\), which is on the \(x\)-axis at \(x =\) 40.0 cm? (b) A point charge \(q = -2.50 \space \mu\)C is placed at \(P\). What are the magnitude and direction of the force exerted by the charge \(q\) \(on\) the ring?

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