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An electric dipole consisting of charges of magnitude \(1.50 \mathrm{n} \mathrm{C}\) separated by \(6.20 \mu \mathrm{m}\) is in an electric tield of strength \(1100 \mathrm{~N} / \mathrm{C} .\) What are (a) the magnitude of the electric dipole moment and (b) the difference between the potential energies for dipole orientations parallel and antiparallel to \(\vec{E} ?\)

Short Answer

Expert verified
(a) \(9.30 \times 10^{-15} \text{ C}\cdot\text{m}\); (b) \(-2.046 \times 10^{-11} \text{ J}\).

Step by step solution

01

Understanding the Problem

We need to find two things: (a) the magnitude of the electric dipole moment, which is a measure of the strength of the dipole; and (b) the potential energy difference when the dipole is aligned parallel and antiparallel to the electric field. The charges have a magnitude of 1.50 nC and they are separated by a distance of 6.20 μm.
02

Calculate the Electric Dipole Moment

The electric dipole moment (\( p \)) is calculated using the formula \( p = q \cdot d \), where \( q \) is the charge and \( d \) is the separation between charges. Substitute \( q = 1.50 \times 10^{-9} \) C and \( d = 6.20 \times 10^{-6} \) m: \[ p = 1.50 \times 10^{-9} \text{ C} \times 6.20 \times 10^{-6} \text{ m} = 9.30 \times 10^{-15} \text{ C}\cdot\text{m} \]
03

Calculate Potential Energy Difference

The potential energy difference for dipole orientations is given by the formula \( \Delta U = -pE \cos(\theta_{\text{parallel}}) + pE \cos(\theta_{\text{antiparallel}}) \). For parallel alignment, \( \theta_{\text{parallel}} = 0^\circ \), and for antiparallel, \( \theta_{\text{antiparallel}} = 180^\circ \). Therefore, \[ \Delta U = -(pE \cos(0)) + (pE \cos(180)) \] where, \( \cos(0) = 1 \) and \( \cos(180) = -1 \), simplifying to:\[ \Delta U = -pE + (-pE) = -2pE \] Substituting \( p = 9.30 \times 10^{-15} \text{ C}\cdot\text{m} \) and \( E = 1100 \text{ N/C} \):\[ \Delta U = -2 \times 9.30 \times 10^{-15} \text{ C}\cdot\text{m} \times 1100 \text{ N/C} = -2.046 \times 10^{-11} \text{ J} \]
04

Finalizing the Solutions

Compile the results: (a) The electric dipole moment is \( 9.30 \times 10^{-15} \text{ C}\cdot\text{m} \).(b) The potential energy difference 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 and opposite charges separated by a distance. The electric dipole moment, denoted as \( p \), represents the strength and orientation of this setup. It is crucial for understanding physical interactions in electric fields.
To compute the dipole moment, we use the equation \( p = q \cdot d \), where:
  • \( q \) is the magnitude of the charge.
  • \( d \) is the distance between the charges.
For example, if the charges are \( 1.50 \mathrm{nC} \) or \( 1.50 \times 10^{-9} \text{ C} \), and they're separated by \( 6.20 \mu \mathrm{m} \) or \( 6.20 \times 10^{-6} \text{ m} \), the dipole moment becomes: \[ p = 1.50 \times 10^{-9} \text{ C} \times 6.20 \times 10^{-6} \text{ m} = 9.30 \times 10^{-15} \text{ C}\cdot\text{m} \]This value provides insight into how strongly the dipole can influence or be influenced by external electric fields.
Potential Energy Difference
The potential energy of a dipole in an electric field depends on its orientation relative to the field. When a dipole changes orientation, there is a difference in potential energy, which is vital for understanding how dipoles behave in fields.
To find the energy difference, we consider the formula for potential energy, \( U = -pE \cos(\theta) \). Here:
  • \( p \) is the dipole moment.
  • \( E \) is the electric field strength.
  • \( \theta \) is the angle between \( p \) and \( E \).
For parallel orientation, \( \theta = 0^\circ \), so \( \cos(0^\circ) = 1 \). For antiparallel, \( \theta = 180^\circ \), so \( \cos(180^\circ) = -1 \). Therefore, the potential energy difference, \( \Delta U \), is:\[ \Delta U = -pE(\cos(0) - \cos(180)) = -pE(1 - (-1)) = -2pE \] Using the given \( p = 9.30 \times 10^{-15} \text{ C}\cdot\text{m} \) and \( E = 1100 \text{ N/C} \), we find:\[ \Delta U = -2 \times 9.30 \times 10^{-15} \text{ C}\cdot\text{m} \times 1100 \text{ N/C} = -2.046 \times 10^{-11} \text{ J} \]This energy change highlights the potential for work when orienting dipoles in fields.
Electric Field Strength
Electric field strength, represented by \( E \), measures how strong an electric field is at a given point. This is crucial for predicting how charges within the field will react.
It's defined mathematically as \( E = F/q \), where \( F \) is the force experienced by a positive test charge \( q \). The units for electric field strength are Newtons per Coulomb (N/C).
A stronger field means that a test charge would experience a greater force. For the dipole considered here, the electric field strength is \( 1100 \text{ N/C} \), indicating a robust field that will significantly influence the dipole’s orientation.
Understanding the electric field is vital because it determines how much work is performed when a dipole aligns or realigns in the field. It also influences the calculation of potential energy differences as shown in the previous section. This foundational concept is key to mastering topics in electromagnetism.

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Most popular questions from this chapter

A particle of charge \(-q_{1}\) is at the origin of an \(x\) axis. (a) At what location on the axis should a particle of charge \(-4 q_{1}\) be placed so that the net electric field is zero at \(x=2.0 \mathrm{~mm}\) on the axis? (b) If, instead, a particle of charge \(+4 q_{1}\) is placed at that location, what is the direction (relative to the positive direction of the \(x\) axis) of the net electric field at \(x=2.0 \mathrm{~mm} ?\)

In Fig. \(22-50,\) a thin glass rod forms a semicircle of radius \(r=5.00 \mathrm{~cm}\). Charge is uniformly distributed along the rod, with \(+q=4.50 \mathrm{pC}\) in the upper half and \(-q=-4.50 \mathrm{pC}\) in the lower half. What are the (a) magnitude and (b) direction (relative to the positive direction of the \(x\) axis) of the electric field \(\vec{E}\) at \(P,\) the center of the semicircle?

Two charged particles are attached to an \(x\) axis: Particle 1 of charge \(-2.00 \times 10^{-7} \mathrm{C}\) is at position \(x=6.00 \mathrm{~cm}\) and particle 2 of charge \(+2.00 \times 10^{-7} \mathrm{C}\) is at position \(x=21.0 \mathrm{~cm} .\) Midway between the particles, what is their net electric field in unit-vector notation?

A clock face has negative point charges \(-q,-2 q,-3 q, \ldots\) \(-12 q\) fixed at the positions of the corresponding numerals. The clock hands do not perturb the net field due to the point charges. At what time does the hour hand point in the same direction as the electric field vector at the center of the dial? (Hint: Use symmetry.)

An electric field \(\vec{E}\) with an average magnitude of about \(150 \mathrm{~N} / \mathrm{C}\) points downward in the atmosphere near Earth's surface. We wish to "float" a sulfur sphere weighing \(4.4 \mathrm{~N}\) in this field by charging the sphere. (a) What charge (both sign and magnitude) must be used? (b) Why is the experiment impractical?

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