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A dipole consisting of charges \(\pm e, 220 \mathrm{nm}\) apart, is placed between two very large (essentially infinite) sheets carrying equal but opposite charge densities of 125\(\mu \mathrm{C} / \mathrm{m}^{2}\) . (a) What is the maximum potential energy this dipole can have due to the sheets, and how should it be oriented relative to the sheets to attain this value? (b) What is the maximum torque the sheets can exert on the dipole, and how should it be oriented relative to the sheets to attain this value? (c) What net force do the two sheets exert on the dipole?

Short Answer

Expert verified
(a) \( U_{max} = -4.35 \times 10^{-20} \text{ J} \), when parallel to field. (b) \( \tau_{max} = 3.03 \times 10^{-25} \text{ N}\cdot\text{m} \), when perpendicular to field. (c) Net force is zero.

Step by step solution

01

Identify Given Values

We are given two sheets with equal and opposite charge densities \( \sigma = 125 \mu C/m^2 \). The dipole consists of charges \( \pm e \) (where \( e \) is the elementary charge) separated by a distance \( d = 220 \text{ nm} \).
02

Calculate Electric Field due to Sheets

A single infinite sheet with surface charge density \( \sigma \) produces an electric field \( E = \frac{\sigma}{2\epsilon_0} \). Two sheets with equal and opposite charge densities produce an electric field \( E = \frac{\sigma}{\epsilon_0} \) between them, directed from the positively charged sheet to the negatively charged sheet.
03

Compute Maximum Potential Energy of Dipole

The potential energy \( U \) of a dipole in an electric field is given by \( U = -\mathbf{p} \cdot \mathbf{E} = -pE\cos(\theta) \), where \( p = ed \) (magnitude of the dipole moment) and \( \theta \) is the angle between \( \mathbf{p} \) and \( \mathbf{E} \). The maximum potential energy occurs when \( \theta = 180^\circ \), so \( U_{max} = edE \). Substitute \( p = ed \) and \( E = \frac{\sigma}{\epsilon_0} \) to get \( U_{max} = -ed \frac{\sigma}{\epsilon_0} \). Calculating with the given values: \( U_{max} = -ed \frac{125 \times 10^{-6} C/m^2}{8.85 \times 10^{-12} C^2/N \cdot m^2} \).
04

Calculate Maximum Torque on Dipole

The torque \( \tau \) on a dipole in a uniform electric field is given by \( \tau = pE\sin(\theta) \). The maximum torque occurs when \( \theta = 90^\circ \), which gives \( \tau_{max} = pE = ed \frac{\sigma}{\epsilon_0} \). Use the given values to calculate: \( \tau_{max} = ed \frac{125 \times 10^{-6}}{8.85 \times 10^{-12}} \).
05

Evaluate Net Force on Dipole

The net force on a dipole in a uniform electric field is zero because the opposite forces on the individual charges cancel each other out.

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Key Concepts

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

Potential Energy
Potential energy relates to the stored energy within a system due to its position in an electric field. When we consider an electric dipole, which consists of two equal and opposite charges separated by a distance, its potential energy depends on how it is oriented in an electric field.

The formula for the potential energy (\( U \)) of a dipole in an electric field (\( E \)) is:
  • \( U = -\mathbf{p} \cdot \mathbf{E} \)
  • \( U = -pE\cos(\theta) \)
where \( \mathbf{p} = ed \) is the dipole moment and \( \theta \) is the angle between the dipole moment and the electric field.

To achieve the maximum potential energy, the angle should be 180 degrees. At this alignment, the equation becomes:
  • \( U_{max} = -ed \frac{\sigma}{\epsilon_0} \)
This setup makes the dipole's stored energy most negative, indicating a strong interaction with the field due to the orientation.
Torque
Torque measures the tendency of a force to rotate an object around an axis. For electric dipoles in fields, this relates to how the dipole aligns with the field. The formula for torque (\( \tau \)) exerted on a dipole by an electric field is:
  • \( \tau = pE\sin(\theta) \)
Here, \( \mathbf{p} \) is the dipole moment, \( E \) is the electric field, and \( \theta \) is the angle between the dipole moment and the field.

The maximum torque occurs when \( \theta = 90^\circ \) (right angle), leading to the largest lever effect:
  • \( \tau_{max} = ed \frac{\sigma}{\epsilon_0} \)
This expression shows how strongly the field can turn the dipole when it is perpendicular to the field direction, maximizing rotational effect.
Electric Field
An electric field describes a region around a charged object where other charged objects experience a force. The strength and direction of this field depend on the charge distribution creating it.

For two infinite sheets with equal and opposite charge densities (\( \sigma \)), the electric field between them is uniform. The magnitude of the electric field (\( E \)) can be calculated using:
  • \( E = \frac{\sigma}{\epsilon_0} \)
Here, \( \epsilon_0 \) represents the permittivity of free space, a constant that determines how electric fields propagate in a vacuum.

This uniform field allows us to compute forces and torques on dipoles within it, making it a common scenario in physics problems involving electric forces.
Charge Density
Charge density refers to the amount of electric charge per unit area on a surface. It is denoted by \( \sigma \) and typically measured in \( \text{C/m}^2 \).

In the context of large sheets, knowing the charge density helps calculate the electric field these sheets produce. Charge density directly impacts the electric field as shown by the formula:
  • \( E = \frac{\sigma}{\epsilon_0} \)
This implies that larger charge densities produce stronger electric fields.

The alignment and movement of a dipole between sheets depend heavily on this measure, influencing potential energy and torque experienced by the dipole.

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Most popular questions from this chapter

A charge of \(-6.50 \mathrm{nC}\) is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25 \(\mathrm{cm}\) . (a) Find the magnitude and direction of the electric field this disk produces at a point \(P\) on the axis of the disk a distance of 2.00 \(\mathrm{cm}\) from its center. (b) Suppose that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. Find the magnitude and direction of the electric field at point \(P\) . (c) If the charge is all brought to the center of the disk, find the magnitude and direction of the electric field at point \(P .\) (d) Why is the field in part (a) stronger than the field in part (b)? Why is the field in part (c) the strongest of the three fields?

Two particles having charges \(q_{1}=0.500 \mathrm{nC}\) and \(q_{2}=8.00 \mathrm{nC}\) are separated by a distance of 1.20 \(\mathrm{m} .\) At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

CP A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 \(\mathrm{cm}\) distant from the first, in a time interval of \(1.50 \times 10^{-6}\) s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

CP A thin disk with a circular hole at its center, called an annulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. P21. 104 ). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z-\) plane, with its center at the origin. For an arbitrary point on the \(x\) -axis (the axis of the annulus), find the magnitude and direction of the electric field \(\vec{E} .\) Consider points both above and below the annulus in Fig. P21. \(104 .\) (c) Show that at points on the \(x\) -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\) -axis (but cannot move off the axis. The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section \(14.2 .\) The annulus is held stationary.)

A very long, straight wire has charge per unit length \(1.50 \times 10^{-10} \mathrm{C} / \mathrm{m} .\) At what distance from the wire is the electric- field magnitude equal to 2.50 \(\mathrm{N} / \mathrm{C}\)?

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