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Ch 25: The Electric Potential
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 25: The Electric PotentialProblem 78
Chapter 25, Problem 78

An electric dipole consists of 1.0 g spheres charged to ±2.0 nC at the ends of a 10-cm-long massless rod. The dipole rotates on a frictionless pivot at its center. The dipole is held perpendicular to a uniform electric field with field strength 1000 V/m, then released. What is the dipole's angular velocity at the instant it is aligned with the electric field?

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Calculate the torque acting on the dipole when it is perpendicular to the electric field. The torque (τ) is given by the formula: τ = pE sin(θ), where p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole and the field. Since the dipole is initially perpendicular to the field, sin(θ) = 1.
Determine the dipole moment (p). The dipole moment is defined as p = qd, where q is the charge on each sphere and d is the separation between the charges. Use q = 2.0 nC and d = 10 cm (convert to meters).
Calculate the potential energy difference (ΔU) of the dipole as it rotates from being perpendicular to the field to being aligned with the field. The potential energy of a dipole in an electric field is given by U = -pE cos(θ). Compute ΔU = U_final - U_initial, where U_initial corresponds to θ = 90° and U_final corresponds to θ = 0°.
Apply the principle of conservation of energy. The initial potential energy of the dipole is converted into rotational kinetic energy as it aligns with the field. The rotational kinetic energy is given by KE = (1/2)Iω², where I is the moment of inertia of the dipole and ω is the angular velocity. Set ΔU = KE to solve for ω.
Determine the moment of inertia (I) of the dipole. Since the dipole consists of two spheres of mass m at a distance of d/2 from the pivot, the moment of inertia is I = 2m(d/2)². Use m = 1.0 g (convert to kg) and d = 10 cm (convert to meters). Substitute I into the energy equation to solve for ω.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Dipole Moment

The electric dipole moment is a vector quantity that represents the separation of positive and negative charges in a dipole. It is calculated as the product of the charge magnitude and the distance between the charges. In this case, the dipole moment influences how the dipole interacts with an external electric field, determining the torque experienced by the dipole when it is misaligned with the field.
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Torque in an Electric Field

When an electric dipole is placed in an electric field, it experiences a torque that tends to align it with the field. The torque ( au) is given by the equation τ = pE sin(θ), where p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field. This torque causes the dipole to rotate, and understanding this relationship is crucial for determining its angular velocity.
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Conservation of Energy

The principle of conservation of energy states that the total energy in a closed system remains constant. In the context of the dipole, the potential energy associated with its orientation in the electric field is converted into kinetic energy as it rotates. By applying this principle, one can relate the initial potential energy of the dipole when perpendicular to the field to its angular velocity when aligned with the field.
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Related Practice
Textbook Question

A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/Lλ(x)=λ_0x/L, where x is measured from the rod's left end. What is λ0λ_0 in terms of Q and L?

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Textbook Question

Two 2.0-mm-diameter beads, C and D, are 10 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 2.0 nC. Bead D has mass 2.0 g and charge −1.0 nC. If the beads are released from rest, what are the speeds vC and vD at the instant the beads collide?

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Textbook Question

FIGURE P25.72 shows a thin rod with charge Q that has been bent into a semicircle of radius R. Find an expression for the electric potential at the center.

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Textbook Question

You are given the equation(s) used to solve a problem. Finish the solution of the problem: (9.0×109Nm2/C2)q₁q₂/0.030m =90×10−6J; q₁+q₂=40nC.

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Textbook Question

Bead A has a mass of 15 g and a charge of −5.0 nC. Bead B has a mass of 25 g and a charge of −10.0 nC. The beads are held 12 cm apart (measured between their centers) and released. What maximum speed is achieved by each bead?

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Textbook Question

The wire in FIGURE P25.74 has linear charge density λ. What is the electric potential at the center of the semicircle?

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