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Chapter 23: Problem 7

Which of the following angles between an electric dipole moment and an applied electric field will result in the most stable state? a) \(0 \mathrm{rad}\) d) The electric dipole moment is b) \(\pi / 2\) rad not stable under any condition in c) \(\pi\) rad an applied electric field.

Short Answer

Expert verified
Answer: a) \(0 \mathrm{rad}\)

Step by step solution

01

Understand the electric dipole moment and potential energy

In a system with an electric dipole moment and an applied electric field, the potential energy (U) of the system can be described as: $$ U = -\vec{p} \cdot \vec{E} $$ where \(\vec{p}\) is the electric dipole moment and \(\vec{E}\) is the applied electric field. The dot product between the two vectors depends on the angle between them, denoted as \(\theta\). Therefore, the potential energy formula can be rewritten as: $$ U = -p E \cos{\theta} $$
02

Analyze the potential energy with respect to the angle θ

We need to determine which angle θ between the electric dipole moment and applied electric field results in the most stable state. The most stable state corresponds to the lowest potential energy. We can examine the given options one by one. Let's evaluate the potential energy for each given angle: 1. Angle θ = \(0 \mathrm{rad}\): $$ U_{0} = -p E \cos{(0 \mathrm{rad})} = -p E $$ 2. Angle θ = \(\pi / 2\) rad: $$ U_{\pi/2} = -p E \cos{(\pi / 2 \mathrm{rad})} = 0 $$ 3. Angle θ = \(\pi\) rad: $$ U_{\pi} = -p E \cos{(\pi \mathrm{rad})} = p E $$
03

Determine the angle corresponding to the most stable state

To find the most stable state, we need to find the angle that results in the lowest potential energy. Comparing the values of the potential energy calculated in step 2, we can see that: - \(U_{0}\) = \(-p E\): The potential energy is negative, indicating a stable state. - \(U_{\pi/2}\) = \(0\): The potential energy is equal to zero, indicating a less stable state than when the potential energy is negative. - \(U_{\pi}\) = \(p E\): The potential energy is positive, indicating an unstable state. Therefore, the angle that results in the most stable state is \(0 \mathrm{rad}\), as it leads to the lowest potential energy. Hence, the correct answer is: a) \(0 \mathrm{rad}\)

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Most popular questions from this chapter

A \(2.50-\mathrm{mg}\) dust particle with a charge of \(1.00 \mu \mathrm{C}\) falls at a point \(x=2.00 \mathrm{~m}\) in a region where the electric potential varies according to \(V(x)=\left(2.00 \mathrm{~V} / \mathrm{m}^{2}\right) x^{2}-\left(3.00 \mathrm{~V} / \mathrm{m}^{3}\right) x^{3}\) With what acceleration will the particle start moving after it touches down?

An electron is accelerated from rest through a potential difference of \(370 \mathrm{~V}\). What is its final speed?

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