/*! 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 59 A tiny electric dipole \(\left(\... [FREE SOLUTION] | 91Ó°ÊÓ

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A tiny electric dipole \(\left(\mathrm{H}_{2} \mathrm{O}\right.\) modecule) of dipole moment \(\vec{p}\) is placed at a distance \(r\) form an infinitely long wire, with its \(\vec{p}\) normal to the wire in the same plane. If the linear charge density of the wire is \(\lambda\), find the electrostatic force acting on the dipole is equal to

Short Answer

Expert verified
The force on the dipole is \[-\frac{\lambda p}{2\pi\varepsilon_0 r^2}.\]

Step by step solution

01

Understanding the Scenario

We have a dipole with a dipole moment \(\vec{p}\) placed at a distance \(r\) from an infinitely long wire. The wire has a linear charge density \(\lambda\), which affects the electric field around it. The problem asks us to find the electrostatic force on the dipole due to this charged wire.
02

Electric Field Due to a Charged Wire

For an infinitely long line of charge with linear charge density \(\lambda\), the electric field \(\vec{E}\) at a distance \(r\) from the wire can be given by the formula: \[E = \frac{\lambda}{2\pi\varepsilon_0 r}\]where \(\varepsilon_0\) is the permittivity of free space. The direction of this field is radially outward from the wire.
03

Components of the Dipole Force

A dipole in an electric field experiences a torque, but we are interested in the force in this exercise. The force on a dipole in a non-uniform electric field is given by the expression:\[ \vec{F} = (\vec{p} \cdot abla) \vec{E}.\]Given the symmetry, we will focus on the gradients that are parallel to the field direction along the line perpendicular to the wire.
04

Force on the Dipole

We need to differentiate the electric field to find the force, utilizing the formula for the force on a dipole:\[F = p \frac{dE}{dr}.\]Taking the derivative of \(E\) with respect to \(r\), we have:\[ \frac{dE}{dr} = -\frac{\lambda}{2\pi\varepsilon_0 r^2}.\]The force is thus:\[F = -\frac{\lambda p}{2\pi\varepsilon_0 r^2}.\]The force acts towards the wire when the dipole is oriented such that attraction occurs due to the configuration.

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Key Concepts

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

Electric Dipole
An electric dipole consists of two equal and opposite charges separated by a small distance. In simpler terms, it's like having a tiny magnet with a positive end and a negative end. The measure of this dipole's strength is given by its dipole moment, represented by the vector \(\vec{p}\). The dipole moment depends on both the amount of charge and the separation distance between the charges. Imagine a water molecule (\(\mathrm{H}_2\mathrm{O}\)) which serves as a classic example of a natural electric dipole, where the molecule's shape and charge distribution create distinct positive and negative sides.

When placed in an external electric field, the electric dipole will experience two things: torque and force. The torque tends to align the dipole with the direction of the field, much like how a compass needle lines up with the Earth's magnetic field. However, in our scenario, we are primarily interested in the force that acts on the dipole due to a charged wire, which is why knowing how the dipole interacts with the electric field is essential.
Linear Charge Density
Linear charge density, denoted as \(\lambda\), is a way to describe how much electric charge is distributed along a line, like a wire. Imagine sprinkling charges evenly along a long, straight wire. The linear charge density tells us how much charge exists per unit length of the wire. It's measured in units of charge per length, such as coulombs per meter (C/m).

This concept is important when calculating the electric field generated by the wire. Since the wire is infinitely long, the total charge would be infinite, so only the charge per meter (\(\lambda\)) is practical to consider. The symmetry of the problem simplifies things because despite the infinite extension, the field calculation only depends on the distance from the wire and the charge per length.
  • Key Formula: Electric Field from a line charge at a distance \(r\), \(E = \frac{\lambda}{2\pi\varepsilon_0 r}\)
Electric Field
The electric field represents the force that a charge would experience in the presence of other electric charges. It's like an invisible force field emanating from charged objects, pulling or pushing on other charges within its reach. For our scenario, we focus on the electric field produced by the infinite wire with linear charge density \(\lambda\).

The electric field at a point a distance \(r\) from this wire is radial, meaning it extends directly outward from the wire. The strength of this field decreases with increasing distance from the wire, following the relation \(E = \frac{\lambda}{2\pi\varepsilon_0 r}\), where \(\varepsilon_0\) stands for the permittivity of free space.

Given this electric field, a dipole feeling its influence will experience a force. The interaction is more sophisticated than with a single charge, as it results from variations in the field acting differently on the dipole's positive and negative ends. Calculating the exact force requires understanding the dipole's unique placement within the field and its orientation. This results in the force expression \(F = -\frac{\lambda p}{2\pi\varepsilon_0 r^2}\), which shows that the dipole is drawn towards the wire under certain orientations.

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Most popular questions from this chapter

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{~N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\), and \(\alpha=30.0^{\circ}\)

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L\). Find the magnitude and direction of the net force on a point charge \(-3 q\) placed (a) at the center of the square and (b) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3 q\) charge by each of the other three charges.

Positive charge \(Q\) is distributed uniformly along the positive \(y\)-axis between \(y=0\) and \(y=a\). A negative point charge \(-q\) lies on the positive \(x\)-axis, a distance \(x\) from the origin (Fig. P1.48). (a) Calculate the \(x\) - and \(y\)-components of the electric field produced by the charge distribution \(Q\) at points on the positive \(x\)-axis. (b) Calculate the \(x\) - and \(y\)-components of the force that the charge distribution \(Q\) exerts on \(q\). (c) Show that if \(x \gg a, F_{x} \cong-Q q / 4 \pi \epsilon_{0} x^{2}\) and \(F_{y} \cong+Q q a / 8 \pi \epsilon_{0} x^{3} .\) Explain why this result is obtained.

Two very large parallel sheets are \(5.00 \mathrm{~cm}\) apart. Sheet A carries a uniform surface charge density of \(-9.50 \mu \mathrm{C} / \mathrm{m}^{2}\), and sheet \(B\), which is to the right of \(A\), carries a uniform charge density of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2}, .\) Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) \(4.00 \mathrm{~cm}\) to the right of sheet \(A ;\) (b) \(4.00 \mathrm{~cm}\) to the left of sheet \(A ;\) (c) \(4.00 \mathrm{~cm}\) to the right of sheet \(B\).of the net force that these wires exert on it?

The earth has a net electric charge that causes a field at points near its surface equal to \(150 \mathrm{~N} / \mathrm{C}\) and directed in toward the center of the earth. (a) What magnitude and sign of charge would a \(60-\mathrm{kg}\) human have to acquire to overcome his or her weight by the force exerted by the earth's electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of \(100 \mathrm{~m}\) ? Is use of the earth's electric field a feasible means of flight? Why or why not?

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