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Chapter 20: Problem 63

A molecule has its dipole moment aligned with a 1.2 -kN/C electric field. If it takes \(3.1 \times 10^{-27} \mathrm{J}\) to reverse the molecule's orientation, what's its dipole moment?

Short Answer

Expert verified
The dipole moment of the molecule is \(1.29 \times 10^{-30} C.m\).

Step by step solution

01

Understand the principle

The work done \(W\) to rotate an electric dipole from an angle \(\theta_1\) to \(\theta_2\) in a uniform electric field \(E\) is given by the formula: \(W = pE(\cos\theta_1 - \cos\theta_2)\), where \(p\) is the electric dipole moment. The problem states that the work done is needed to reverse the orientation of the dipole. This means that the dipole is flipped 180 degrees, or in other words, \(\theta_2 - \theta_1 = 180^o\) or \(\theta_1 = 0\) and \(\theta_2 = \pi\). Reversing the dipole means that the initial and final orientations are aligned and anti-aligned with the electric field respectively. So \(\cos\theta_1 = 1\) and \(\cos\theta_2 = -1\). So, the formula simplifies to \(W = 2pE\).
02

Substitute the values

Substitute the given values into the formula \(W = 2pE\). We know that \(W = 3.1 \times 10^{-27} J\) (work done to reverse the orientation) and \(E = 1.2 kN/C = 1.2 \times 10^3 N/C\) (electric field). Solving for \(p\), we get: \(p = \frac{W}{2E}\).
03

Calculate the dipole moment

Substitute \(W = 3.1 \times 10^{-27} J\) and \(E = 1.2 \times 10^3 N/C\) into the formula to get: \(p = \frac{3.1 \times 10^{-27} J}{2 \times 1.2 \times 10^3 N/C} = 1.29 \times 10^{-30} C.m\).

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