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Chapter 23: Problem 52

Dipole in a Field An electric dipole, consisting of charges of magnitude \(1.50 \mathrm{nC}\) separated by \(6.20 \mu \mathrm{m}\), is in an electric field of strength \(1100 \mathrm{~N} / \mathrm{C}\). (a) What is the magnitude of the electric dipole moment? (b) What is the difference between the potential energies corresponding to dipole orientations parallel to and antiparallel to the field?

Short Answer

Expert verified
(a) The electric dipole moment is \( 9.3 \times 10^{-15} \text{ C} \text{ m} \). (b) The potential energy difference is \( 2.046 \times 10^{-11} \text{ J} \).

Step by step solution

01

Calculate the Electric Dipole Moment

The electric dipole moment (\textbf{p}) can be calculated using the formula: \[ p = q \times d \] where \( q \) is the charge magnitude ( \( 1.50 \times 10^{-9} \) C) \( d \) is the separation distance ( \( 6.20 \times 10^{-6} \) m). Substitute the given values: \[ p = (1.50 \times 10^{-9} \text{ C}) \times (6.20 \times 10^{-6} \text{ m}) = 9.30 \times 10^{-15} \text{ C} \text{ m} \] Therefore, the electric dipole moment is \( p = 9.30 \times 10^{-15} \text{ C} \text{ m} \).
02

Calculate the Potential Energy Difference

The potential energy ( \( U \) ) of a dipole in an electric field can be calculated using: \[ U = - p E \times \text{cos}(\theta) \] We need to find the potential energy difference ( \( \text{螖}U \) ) between parallel ( \( \theta = 0^{\boxed{\tiny \text{o}}} \) ) and antiparallel ( \( \theta = 180^{\boxed{\tiny \text{o}}} \) ) orientations: \[ \text{螖}U = U_{\text{antiparallel}} - U_{\text{parallel}} \] Substitute \( p = 9.30 \times 10^{-15} \text{ C} \text{ m} \) and \( E = 1100 \text{ N} / \text{ C} \): - For parallel ( \( \theta = 0^{\boxed{\tiny \text{o}}} \) ): \[ U_{\text{parallel}} = - p E \text{cos}(0^{\boxed{\tiny \text{o}}}) = - (9.30 \times 10^{-15} \text{ C} \text{ m}) \times (1100 \text{ N}/\text{C}) \times 1 = - 1.023 \times 10^{-11} \text{ J} \] - For antiparallel ( \( \theta = 180^{\boxed{\tiny \text{o}}} \) ): \[ U_{\text{antiparallel}} = - p E \text{cos}(180^{\boxed{\tiny \text{o}}}) = - (9.30 \times 10^{-15} \text{ C} \text{ m}) \times (1100 \text{ N}/\text{C}) \times (-1) = 1.023 \times 10^{-11} \text{ J} \] Now, find the difference: \[ \text{螖}U = 1.023 \times 10^{-11} \text{ J} - ( - 1.023 \times 10^{-11} \text{ J}) = 2.046 \times 10^{-11} \text{ J} \] Therefore, the difference in potential energies is \( 2.046 \times 10^{-11} \text{ J} \).

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Key Concepts

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

electric dipole moment
An electric dipole consists of two equal but opposite charges separated by a small distance. The strength of this dipole is described by the electric dipole moment. This is a vector quantity denoted by \textbf{p}, which points from the negative to the positive charge.
The formula to calculate the electric dipole moment is:
\[ p = q \times d \]
Here, \(q\) is the magnitude of either charge, and \(d\) is the distance between them.
For example:
  • Suppose the charges are \(1.50 \text{ nC} = 1.50 \times 10^{-9} \text{ C}\)
  • The distance separating them is \(6.20 \text{ 渭m} = 6.20 \times 10^{-6} \text{ m}\)

Then the dipole moment is \(p = (1.50 \times 10^{-9} \text{ C}) \times (6.20 \times 10^{-6} \text{ m}) = 9.30 \times 10^{-15} \text{ C} \text{m}\).
This tiny product gives us the measure of how strong and in which direction the dipole's effect will be felt in an electric field.
potential energy in electric fields
The potential energy of an electric dipole in a uniform electric field depends on its orientation relative to the field.
It is given by the formula:
\[ U = - p E \times \text{cos}(\theta) \]
Where:
  • \(U\) is the potential energy
  • \(p\) is the dipole moment
  • \(E\) is the electric field strength
  • \(\theta\) is the angle between the dipole moment and the electric field.
When the dipole is aligned parallel to the field (\(\theta = 0^{\boxed{\text{o}}}\)), the potential energy is minimized:
\[ U_{\text{parallel}} = - p E \times \text{cos}(0^{\boxed{\text{o}}}) = - p E \]
In the opposite case, where the dipole is aligned antiparallel (\(\theta = 180^{\boxed{\text{o}}}\)), the potential energy is maximized:
\[ U_{\text{antiparallel}} = - p E \times \text{cos}(180^{\boxed{\text{o}}}) = p E \]
The difference between these energies is:
\[ \text{螖}U = U_{\text{antiparallel}} - U_{\text{parallel}} = p E - (- p E) = 2 p E \]
In our example, with \(p = 9.30 \times 10^{-15} \text{ C} \text{m}\) and \(E = 1100 \text{ N} / \text{C}\), we get:
\[ \text{螖}U = 2 \times (9.30 \times 10^{-15} \text{ C} \text{m}) \times 1100 \text{ N} / \text{C} = 2.046 \times 10^{-11} \text{ J} \]
This energy difference indicates how much work is needed to rotate the dipole 180 degrees within the electric field.
dipole orientation
The orientation of a dipole in an electric field is crucial as it affects the dipole's potential energy and behavior.
When positioned in an electric field, there are three primary orientations:
  • Parallel: The dipole is aligned with the field direction (\(\theta = 0^{\boxed{\text{o}}}\)). This orientation has the lowest potential energy as it is the most stable alignment.
  • Antiparallel: The dipole is aligned opposite to the field direction (\(\theta = 180^{\boxed{\text{o}}}\)). This orientation has the highest potential energy, reflecting instability and a tendency to reorient to parallel.
  • Perpendicular: The dipole is at a right angle to the field (\(\theta = 90^{\boxed{\text{o}}}\)). Here, the potential energy is neutral or zero because the dipole is not aligned with the field to lower or heighten the energy.
Such orientations are essential in analyzing how dipoles react within diverse electric fields.
Generally, a dipole will naturally move from higher potential energy to a lower one; from antiparallel to parallel. This concept helps in understanding molecular configurations and reactions in electric fields, predicting the behavior of complex systems, and even describing phenomena in chemistry and physics.
Knowing these orientations aids in visualizing how forces act on dipoles in fields, providing clarity on real-world applications and theoretical problems.

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Most popular questions from this chapter

Two Curved Plastic Rods In Fig. \(23-40 b\), two curved plastic rods, one of charge \(+q\) and the other of charge \(-q\) form a circle of radius \(R\) in an \(x y\) plane. The \(x\) axis passes through their connecting points, and the charge is distributed uniformaly on both rods. What are the magnitude and direction of the electric field \(\vec{E}\) produced at \(P\), the center of the circle?

Thin Circular Disk Sketch qualitatively the electric field lines for a thin, circular, uniformly charged disk of radius \(R\). (Hint: Consider as limiting cases points very close to the disk, where the electric field is directed perpendicular to the surface, and points very far from it, where the electric field is like that of a point charge.)

Humid Air Humid air breaks down (its molecules become ionized) in an electric field of \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\). In that field, what is the magnitude of the electrostatic force on (a) an electron and (b) an ion with a single electron missing?

Entering a Field An electron enters a region of uniform electric field with an initial velocity of \(40 \mathrm{~km} / \mathrm{s}\) in the same direction as the electric field, which has magnitude \(E=50 \mathrm{~N} / \mathrm{C}\). (a) What is the speed of the electron \(1.5\) ns after entering this region? (b) How far does the electron travel during the \(1.5 \mathrm{~ns}\) interval?

Uniform Upward Field In Fig. \(23-49\), a uniform, upwarddirected electric field \(\vec{E}\) of magnitude \(2.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\) has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length \(L=10.0 \mathrm{~cm}\) and separation \(d=2.00 \mathrm{~cm} .\) An electron is then shot between the plates from the left edge of the lower plate. The initial velocity \(\vec{v}_{1}\) of the electron makes an angle \(\theta=45.0^{\circ}\) with the lower plate and has a magnitude of \(6.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). (a) Will the electron strike one of the plates? (b) If so, which plate and how far horizontally from the left edge will the electron strike?
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