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Chapter 21: Problem 101

In a follow-up experiment, a charge of \(+40 \mathrm{pC}\) was placed at the center of an artificial flower at the end of a \(30-\mathrm{cm}\) -long stem. Bees were observed to approach no closer than \(15 \mathrm{~cm}\) 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) \(2.4 \mathrm{~N} / \mathrm{C} ;\) (b) \(16 \mathrm{~N} / \mathrm{C} ;\) (c) \(2.7 \times 10^{-10} \mathrm{~N} / \mathrm{C}\) (d) \(4.8 \times 10^{-10} \mathrm{~N} / \mathrm{C}\).

Short Answer

Expert verified
The smallest external electric field to which bees may be sensitive is closest to \(2.4 \mathrm{N/C}\)

Step by step solution

01

Convert values to standard units

First, convert the given values to standard units. The charge \(Q = 40 pC = 40 \times 10^{-12} C\) and the distance \(r = 15 cm = 0.15 m\).
02

Apply the formula for electric field

The formula for the electric field is \(E = k \frac{Q}{r^2}\), where \(k = 8.99 \times 10^{9} N \cdot m^2 / C^2\). Note that the denominator \(r^2\) is the square of the distance from the charge.
03

Calculate the value of the electric field

Substitute the given values into the formula: \(E = 8.99 \times 10^{9} N \cdot m^2 / C^2 \times \frac{40 \times 10^{-12} C}{(0.15 m)^2}\). Work out the calculation to get the magnitude of the electric field.
04

Compare with the given options

Once you have computed the electric field, compare your answer with the options provided in the question. The closest value to your computed answer is the correct choice.

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Key Concepts

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

Electric Field Formula
Understanding the electric field formula is essential when studying the effects of charged particles, like in the case of bees' sensitivity to electric fields. The electric field (\(E\)) represents the force per unit charge exerted on a small positive test charge placed in the vicinity of a source charge.

The standard formula to calculate the electric field due to a point charge is given as: \[ E = k \frac{Q}{r^2} \] where \(E\) is the electric field strength, \(k\) is Coulomb's constant which is \(8.99 \times 10^9 N \cdot m^2/C^2\), \(Q\) is the charge of the source, and \(r\) is the distance from the charge to the point at which the electric field is being measured.

It's important to note when we talk about the 'strength' of an electric field, we are discussing how much force a charge will experience in a given region, which can be a determinant for phenomena like the sensitivity of bees to electric fields.
Coulomb's Law
Coulomb's law is the cornerstone of electrostatics, named after Charles-Augustin de Coulomb, who first described the law in 1785. The law states that the electric force (\(F\)) between two point charges is directly proportional to the product of the magnitudes of the charges (\(Q1\) and \(Q2\)) and inversely proportional to the square of the distance (\(r\)) between them. The formula for Coulomb's law is expressed as: \[ F = k \frac{Q1 \cdot Q2}{r^2} \]

Here, Coulomb's constant \(k\) appears again, serving a similar purpose to provide consistency with the electric field formula. Understanding Coulomb's law is vital because it lays the foundation for the electric field concept, where the electric field is seen as an extension of the force exerted by one charge onto another.
Conversion of Units
Conversion of units in physics is a critical skill, as it ensures that equations are dimensionally consistent and the resulting calculations make practical sense. For instance, in the exercise about bee sensitivity, we convert picocoulombs (\(pC\)) to coulombs (\(C\)) and centimeters to meters.

This conversion is done because the standard unit of charge in the International System of Units (SI) is the coulomb, and the standard unit for distance is meters. Failing to convert to these standard units could result in errors in calculation and incorrect interpretations of the results. It is a straightforward process where we use conversion factors, such as: \[1 pC = 10^{-12} C\] and \[1 cm = 0.01 m\] But despite its simplicity, this step is crucial for the accuracy of all subsequent calculations.
Standard Units in Physics
Standard units in physics are universally accepted measures that enable scientists and engineers from around the globe to communicate and compare results without confusion. The International System of Units (SI), is the most widely used system.

In the context of electric fields and forces, the standard units are:
  • Electric field (E): Newtons per coulomb (N/C or V/m)
  • Charge (Q): Coulombs (C)
  • Distance (r): Meters (m)
  • Force (F): Newtons (N)
By using these established units, we can consistently apply formulas like the electric field formula and Coulomb's law, and ensure that our results are not only correct but also meaningfully comparable in the scientific community.

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Most popular questions from this chapter

Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by \(3.1 \mathrm{~mm}\), forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(36.9^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m} ?\)

If two electrons are each \(1.50 \times 10^{-10} \mathrm{~m}\) from a proton (Fig. E21.41), find the magnitude and direction of the net electric force they will exert on the proton.

Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of \(1.4 \times 10^{-8} \mathrm{~g}\). The drops will leave the nozzle and travel toward the paper at \(50 \mathrm{~m} / \mathrm{s}\) in a horizontal direction, passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops will then pass between parallel deflecting plates, \(2.0 \mathrm{~cm}\) long, where there is a uniform vertical electric field with magnitude \(8.0 \times 10^{4} \mathrm{~N} / \mathrm{C}\). Your team is working on the design of the charging unit that places the charge on the drops. (a) If a drop is to be deflected \(0.30 \mathrm{~mm}\) by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge? (b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of \(25 \mathrm{~m} / \mathrm{s},\) what \(q\) value is needed to achieve the same \(0.30 \mathrm{~mm}\) deflection?

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 +2.00 nC point charge is at the origin, and a second \(-5.00 \mathrm{nC}\) point charge is on the \(x\) -axis at \(x=0.800 \mathrm{~m}\). (a) Find the electric field (magnitude and direction) at each of the following points on the \(x\) -axis: (i) \(x=0.200 \mathrm{~m} ;\) (ii) \(x=1.20 \mathrm{~m} ;\) (iii) \(x=-0.200 \mathrm{~m}\). (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).
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