/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A small rigid object carries pos... [FREE SOLUTION] | 91Ó°ÊÓ

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A small rigid object carries positive and negative 3.50 -nC charges. It is oriented so that the positive charge has coordinates \((-1.20 \mathrm{mm}, 1.10 \mathrm{mm})\) and the negative charge is at the point \((1.40 \mathrm{mm},-1.30 \mathrm{mm} \text { ). (a) Find the electric }\) dipole moment of the object. The object is placed in an electric field \(\mathbf{E}=(7800 \hat{\mathbf{i}}-4900 \hat{\mathbf{j}}) \mathrm{N} / \mathrm{C}\) . (b) Find the torque acting on the object. (C) Find the potential energy of the object- field system when the object is in this orientation. (d) If the orientation of the object can change, find the difference between the maximum and minimum potential energies of the system.

Short Answer

Expert verified
The electric dipole moment is parallel to the E field and its magnitude is the charge multiplied by the displacement between charges. The torque is the cross product of the dipole moment and the electric field. The potential energy is the negative dot product of the dipole moment and the electric field. The difference between maximum and minimum potential energies is twice the magnitude of the potential energy in the given orientation.

Step by step solution

01

Calculate the Dipole Moment

The electric dipole moment \(\mathbf{p}\) is given by \(\mathbf{p} = q \mathbf{d}\), where \(q\) is the charge and \(\mathbf{d}\) is the displacement vector pointing from the negative to the positive charge. The charges are \(+3.50 \text{nC}\) and \(-3.50 \text{nC}\), and their positions are provided. Convert the charges to Coulombs by using \(1 \text{nC} = 1 \times 10^{-9} \text{C}\). Then, calculate \(\mathbf{d}\) using the coordinates given and find \(\mathbf{p}\).
02

Find the Torque

The torque \(\mathbf{\tau}\) acting on an electric dipole in a uniform electric field is given by \(\mathbf{\tau} = \mathbf{p} \times \mathbf{E}\), where \(\mathbf{E}\) is the electric field and \(\times\) denotes the cross product. Use the dipole moment calculated from Step 1 and the given electric field to compute the torque.
03

Calculate the Potential Energy

The potential energy \(U\) of a dipole in an electric field is given by \(U = -\mathbf{p} \cdot \mathbf{E}\), where \(\cdot\) denotes the dot product. Using the dipole moment from Step 1 and the electric field, compute the potential energy of the dipole when it is in the given orientation.
04

Evaluate the Potential Energy Extrema

The potential energy \(U\) has a maximum when the dipole moment is antiparallel to \(\mathbf{E}\) and a minimum when \(\mathbf{p}\) is parallel to \(\mathbf{E}\). Calculate the maximum \(U_{\text{max}} = -pE\) and minimum \(U_{\text{min}} = pE\) potential energies, and find the difference by subtracting \(U_{\text{min}}\) from \(U_{\text{max}}\).

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Key Concepts

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

Torque on Dipole in Electric Field
Understanding the torque on an electric dipole in an electric field is crucial in grasping how dipoles behave under the influence of external forces. When an electric dipole, which consists of two opposite charges separated by a distance, is placed in an electric field, it experiences a torque. This torque tends to align the dipole with the electric field. The formula for the torque is given by the cross product of the electric dipole moment (\textbf{p}) and the electric field (\textbf{E}), represented mathematically as \( \mathbf{\tau} = \mathbf{p} \times \mathbf{E} \).

The dipole moment, \textbf{p}, is a vector quantity with its direction pointing from the negative to the positive charge, and its magnitude is the product of one of the charges and the distance between them. The resulting torque vector, \( \mathbf{\tau} \) is perpendicular to the plane formed by \textbf{p} and \textbf{E}. It's important for students to visualize this three-dimensional interaction to fully comprehend the direction and magnitude of the resulting torque.
Potential Energy of a Dipole
The potential energy of a dipole in an electric field is an expression of the work done by the field on the dipole to bring it to a certain orientation from a reference orientation. Typically, this reference is when the dipole is perpendicular to the electric field lines, where it has zero potential energy. The formula for the potential energy (U) of a dipole in an electric field is \( U = -\mathbf{p} \cdot \mathbf{E} \), where \(-\cdot\) indicates the dot product between the electric dipole moment \textbf{p} and the electric field \textbf{E}.

A key aspect to remember is that the potential energy is negative when the dipole is parallel to the electric field, indicating a stable equilibrium. Conversely, it is positive when the dipole is antiparallel to the field, signifying an unstable equilibrium. For students, visualizing the orientation of the dipole relative to the field and understanding that it prefers to align with the field due to lower potential energy can be critical for internalizing this concept.
Uniform Electric Field
A uniform electric field is characterized by electric field lines that are parallel to each other and equally spaced. This implies that the electric field has the same magnitude and direction at every point within the region it occupies. Depicting this concept accurately can help students understand the forces exerted on charges in such fields. The importance of a uniform electric field in the context of an electric dipole is that it provides a consistent external influence on the dipole, resulting in a predictable behavior—like producing a constant torque and a definite potential energy based on the dipole's orientation. This uniformity is what allows us to apply the torque and potential energy formulas reliably. For visual learners, drawing field lines and indicating the vector nature of a uniform electric field can aid comprehension of these interactions.

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Most popular questions from this chapter

An isolated charged conducting sphere of radius 12.0 \(\mathrm{cm}\) creates an electric field of \(4.90 \times 10^{4} \mathrm{N} / \mathrm{C}\) at a distance 21.0 \(\mathrm{cm}\) from its center. (a) What is its surface charge density? (b) What is its capacitance?

Capacitors \(C_{1}=6.00 \mu \mathrm{F}\) and \(C_{2}=2.00 \mu \mathrm{F}\) are charged as a parallel combination across a \(250-\mathrm{V}\) battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.

An air-filled capacitor consists of two parallel plates, each with an area of \(7.60 \mathrm{cm}^{2},\) separated by a distance of \(1.80 \mathrm{mm} .\) A \(20.0-\mathrm{V}\) potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

A 10.0 - \(\mu\) F capacitor has plates with vacuum between them. Each plate carries a charge of magnitude 1000\(\mu \mathrm{C}\) . A particle with charge \(-3.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-16} \mathrm{kg}\) is fired from the positive plate toward the negative plate with an initial speed of \(2.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) . Does it reach the negative plate? If so, find its impact speed. If not, what fraction of the way across the capacitor does it travel?

The immediate cause of many deaths is ventricular fibrillation, uncoordinated quivering of the heart as opposed to proper beating. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart will sometimes start organized beating again. A defibrilator (Fig. 26.14\()\) is a device that applies a strong electric shock to the chest over a time interval of a few milliseconds. The device contains a capacitor of several microfarads, charged to several thousand volts. Electrodes called paddles, about 8 \(\mathrm{cm}\) across and coated with conducting paste, are held against the chest on both sides of the heart. Their handles are insulated to prevent injury to the operator, who calls, "Clear!" and pushes a button on one paddle to discharge the capacitor through the patient's chest. Assume that an energy of 300 \(\mathrm{J}\) is to be delivered from a \(30.0-\mu \mathrm{F}\) capacitor. To what potential difference must it be charged?

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