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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 body of mass \(M\), carrying charge \(Q\), falls from rest from a height \(h\) (above the ground) near the surface of the Earth, where the gravitational acceleration is \(g\) and there is an electric field with a constant component \(E\) in the vertical direction. a) Find an expression for the speed, \(v,\) of the body when it reaches the ground, in terms of \(M, Q, h, g,\) and \(E\). b) The expression from part (a) is not meaningful for certain values of \(M, g, Q,\) and \(E\). Explain what happens in such cases.

A uniformly charged rod of length \(L\) with total charge \(Q\) lies along the \(y\) -axis, from \(y=0\) to \(y=L\). Find an expression for the electric field at the point \((d, 0)\) (that is, the point at \(x=d\) on the \(x\) -axis).

Three charges are on the \(y\) -axis. Two of the charges, each \(-q,\) are located \(y=\pm d,\) and the third charge, \(+2 q,\) is located at \(y=0 .\) Derive an expression for the electric field at a point \(P\) on the \(x\) -axis.

A solid, nonconducting sphere of radius \(a\) has total charge \(Q\) and a uniform charge distribution. Using Gauss's Law, determine the electric field (as a vector) in the regions \(ra\) in terms of \(Q\).

\( \mathrm{~A}+48.00-\mathrm{nC}\) point charge is placed on the \(x\) -axis at \(x=4.000 \mathrm{~m},\) and \(\mathrm{a}-24.00-\mathrm{n} \mathrm{C}\) point charge is placed on the \(y\) -axis at \(y=-6.000 \mathrm{~m} .\) What is the direction of the electric field at the origin?
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