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Chapter 22: Problem 40

Electric dipole moments of molecules are often measured in debyes \((\mathrm{D}),\) where \(1 \mathrm{D}=3.34 \cdot 10^{-30} \mathrm{C} \mathrm{m} .\) For instance, the dipole moment of hydrogen chloride gas molecules is \(1.05 \mathrm{D}\). Calculate the maximum torque such a molecule can experience in the presence of an electric field of magnitude \(160.0 \mathrm{~N} / \mathrm{C}\).

Short Answer

Expert verified
Answer: The maximum torque experienced by a hydrogen chloride gas molecule in the presence of the given electric field is approximately \(5.60 \times 10^{-28} \mathrm{N} \cdot \mathrm{m}\).

Step by step solution

01

Convert dipole moment from debyes to Cm

First, let's convert the given dipole moment from debyes \((D)\) to Coulombs per meter \((Cm)\). Use the conversion factor \(1 \mathrm{D} = 3.34 \cdot 10^{-30} \mathrm{Cm}\): $$ \text{dipole moment in Cm} = 1.05\, \mathrm{D} \times \frac{3.34 \cdot 10^{-30} \mathrm{Cm}}{1\,\mathrm{D}} $$ Calculate the value: $$ \text{dipole moment in Cm} = 1.05 \times 3.34 \times 10^{-30} \mathrm{Cm} \approx 3.50 \times 10^{-30} \mathrm{Cm} $$
02

Find the equation for torque on a dipole

The torque \(\tau\) exerted on an electric dipole in an electric field is given by: $$ \tau = p \cdot E \cdot \sin{\theta} $$ Where \(p\) is the dipole moment, \(E\) is the electric field, and \(\theta\) is the angle between the dipole moment and the electric field direction.
03

Find the maximum torque

To find the maximum torque, we need to maximize the value of \(\sin{\theta}\). \(\sin{\theta}\) is maximized when \(\theta = 90^\circ\), in which case \(\sin{\theta} = 1\): $$ \tau_\text{max} = p \cdot E $$ Substitute the dipole moment (in Cm) and the electric field (in N/C) into the equation: $$ \tau_\text{max} = (3.50 \times 10^{-30} \mathrm{Cm}) \cdot (160.0 \mathrm{N/C}) $$ Calculate the maximum torque: $$ \tau_\text{max} \approx 5.60 \times 10^{-28} \mathrm{N} \cdot \mathrm{m} $$ So, the maximum torque experienced by a hydrogen chloride gas molecule in the presence of the given electric field is approximately \(5.60 \times 10^{-28} \mathrm{N} \cdot \mathrm{m}\).

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Most popular questions from this chapter

A sphere centered at the origin has a volume charge distribution of \(120 \mathrm{nC} / \mathrm{cm}^{3}\) and a radius of \(12 \mathrm{~cm}\). The sphere is centered inside a conducting spherical shell with an inner radius of \(30.0 \mathrm{~cm}\) and an outer radius of \(50.0 \mathrm{~cm}\). The charge on the spherical shell is \(-2.0 \mathrm{mC}\). What is the magnitude and direction of the electric field at each of the following distances from the origin? a) at \(r=10.0 \mathrm{~cm}\) c) at \(r=40.0 \mathrm{~cm}\) b) at \(r=20.0 \mathrm{~cm}\) d) at \(r=80.0 \mathrm{~cm}\)

A point charge, \(+Q\), is located on the \(x\) -axis at \(x=a\), and a second point charge, \(-Q\), is located on the \(x\) -axis at \(x=-a\). A Gaussian surface with radius \(r=2 a\) is centered at the origin. The flux through this Gaussian surface is a) zero. b) greater than zero. c) less than zero. d) none of the above.

There is a uniform charge distribution of \(\lambda=\) \(8.00 \cdot 10^{-8} \mathrm{C} / \mathrm{m}\) along a thin wire of length \(L=6.00 \mathrm{~cm}\) The wire is then curved into a semicircle that is centered about the origin, so the radius of the semicircle is \(R=L / \pi .\) Find the magnitude of the electric field at the center of the semicircle.

Why do electric field lines never cross?

Two infinite sheets of charge are separated by \(10.0 \mathrm{~cm}\) as shown in the figure. Sheet 1 has a surface charge distribution of \(\sigma_{1}=3.00 \mu \mathrm{C} / \mathrm{m}^{2}\) and sheet 2 has a surface charge distribution of \(\sigma_{2}=-5.00 \mu \mathrm{C} / \mathrm{m}^{2}\). Find the total electric field (magnitude and direction) at each of the following locations: a) at point \(P, 6.00 \mathrm{~cm}\) to the left of sheet 1 b) at point \(P^{\prime} 6.00 \mathrm{~cm}\) to the right of sheet 1
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