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Torque on a Dipole An electric dipole consists of charges \(+2 e\) and \(-2 e\) separated by \(0.78 \mathrm{~nm}\). It is in an electric field of strength \(3.4 \times 10^{6} \mathrm{~N} / \mathrm{C}\). Calculate the magnitude of the torque on the dipole when the dipole moment is (a) parallel to, (b) perpendicular to, and (c) antiparallel to the electric field.

Short Answer

Expert verified
(a) 0, (b) 8.4864 脳 10鈦宦猜 Nm, (c) 0

Step by step solution

01

- Understand the Dipole Moment

The electric dipole moment (\textbf{p}) is given by the product of the charge magnitude and the separation distance between the charges. Therefore, \textbf{p} can be calculated as:\[ p = q \times d \] where \( q = 2e = 2 \times 1.6 \times 10^{-19} \text{ C} \) and \( d = 0.78 \text{ nm} = 0.78 \times 10^{-9} \text{ m} \). Substituting the values, we get: \[ p = (2 \times 1.6 \times 10^{-19} \text{ C}) \times (0.78 \times 10^{-9} \text{ m}) = 2.496 \times 10^{-28} \text{ Cm} \]
02

- Torque Formula

The magnitude of the torque (\textbf{\tau}) on a dipole in an electric field is given by:\[ \tau = pE \times \text{sin}(\theta) \] where \( p \) is the dipole moment, \( E \) is the electric field strength, and \( \theta \) is the angle between the dipole moment and the field.
03

- Case (a): Dipole Parallel to Electric Field

When the dipole is parallel to the electric field, \( \theta = 0^\text{o} \):\[ \tau = pE \times \text{sin}(0^\text{o}) = pE \times 0 = 0 \] Thus, the torque is zero.
04

- Case (b): Dipole Perpendicular to Electric Field

When the dipole is perpendicular to the electric field, \( \theta = 90^\text{o} \):\[ \tau = pE \times \text{sin}(90^\text{o}) = pE \times 1 = pE \] Substituting the values, we get:\[ \tau = (2.496 \times 10^{-28} \text{ Cm}) \times (3.4 \times 10^{6} \text{ N/C}) = 8.4864 \times 10^{-22} \text{ Nm} \]
05

- Case (c): Dipole Antiparallel to Electric Field

When the dipole is antiparallel to the electric field, \( \theta = 180^\text{o} \):\[ \tau = pE \times \text{sin}(180^\text{o}) = pE \times 0 = 0 \] Thus, the torque is again zero.

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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 by \(\mathbf{p}\), is a vector quantity. It points from the negative charge to the positive charge, and its magnitude is given by:
  • \( p = q \cdot d \)
where \( q \) is the magnitude of either charge, and \( d \) is the separation distance between the charges. In our example, the charges are \(+2 e\) and \(-2 e\), and they are separated by \(0.78 \text{ nm}\). We convert this to meters for consistency in SI units: \(0.78 \text{ nm} = 0.78 \cdot 10^{-9} \text{ m} \). With \(q\) being \(2 \cdot 1.6 \cdot 10^{-19} \text{ C} \), we get the dipole moment to be \( p = 2.496 \cdot 10^{-28} \text{ Cm} \). This dipole moment is fundamental for understanding the interaction between the dipole and an external electric field.
torque calculation
Torque (\textbf{\tau\}) on a dipole in an electric field describes the rotational effect of the field on the dipole. It depends on the electric field strength (\textbf{E}), the dipole moment (\textbf{p}), and the angle (\textbf{\theta\}) between them. The formula to calculate this is:
  • \( \tau = pE \cdot \sin(\theta) \)
This equation shows that \(\tau\) achieves its maximum value when the dipole is perpendicular to the electric field (\textbf{\theta = 90^\text{o}\}) and drops to zero when the dipole is parallel (\textbf{\theta = 0^\text{o}\}) or antiparallel (\textbf{\theta = 180^\text{o}\}) to the field. Calculating torque helps us understand how a dipole placed in an electric field will rotate, aligning itself with the field direction.
electric field strength
The electric field strength (\textbf{E}) is defined as the force per unit charge exerted by a field on a test charge. It is measured in \( \text{N/C} \) (Newtons per Coulomb). In this exercise, the electric field strength is given as \(3.4 \cdot 10^6 \text{ N/C}\). This strong field will exert significant forces on any charged particles or dipoles within it. Understanding the electric field strength is crucial because it directly influences the magnitude of the torque on the dipole through the equation \( \tau = pE \cdot \sin(\theta) \). A higher electric field strength will yield a larger torque, affecting the rotational dynamics of the dipole.
angle dependence of torque
The angle (\textbf{\theta\}) between the dipole moment (\textbf{p}) and the electric field (\textbf{E}) is crucial in determining the magnitude of the torque. This relationship is given by the sine function in the torque formula: \(\tau = pE \cdot \sin(\theta) \). Here's how different angles affect the torque:
  • Parallel (\theta = 0^\text{o}): \sin(0^\text{o}) = 0\, thus, \(\tau = 0\).
  • Perpendicular (\theta = 90^\text{o}): \sin(90^\text{o}) = 1\, thus, \(\tau = pE\).
  • Antiparallel (\theta = 180^\text{o}): \sin(180^\text{o}) = 0\, thus, \(\tau = 0\).
The torque is greatest when the dipole is perpendicular to the field because \(\sin(90^\text{o}) = 1\). It is zero when the dipole is either parallel or antiparallel to the field because \(\sin(0^\text{o}) \text{and} \sin(180^\text{o}) = 0\). This demonstrates how torque varies with the dipole's orientation relative to the electric field.

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Most popular questions from this chapter

Density, Density, Density. (a) A charge of \(-300 e\) is uniformly distributed along a circular arc of radius \(4.00 \mathrm{~cm}\), which subtends an angle of \(40^{\circ} .\) What is the linear charge density along the arc? (b) A charge of \(-300 e\) is uniformly distributed over one face of a circular disk of radius \(2.00 \mathrm{~cm}\). What is the surface charge density over that face? (c) A charge of \(-300 e\) is uniformly distributed over the surface of a sphere of radius \(2.00 \mathrm{~cm} .\) What is the surface charge density over that surface? (d) A charge of \(-300 e\) is uniformly spread through the volume of a sphere of radius \(2.00 \mathrm{~cm} .\) What is the volume charge density in that sphere?

High-Speed Protons Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were \(2.00 \times 10^{4} \mathrm{~N} / \mathrm{C} ?(\mathrm{~b})\) What speed would the proton attain if the field accelerated the proton through a distance of \(1.00 \mathrm{~cm}\) ?

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