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Chapter 1: Problem 62

A thin non-conducting ring of radius \(R\) has a linear charge density \(\lambda=\lambda_{0} \cos \theta\), where \(\lambda_{0}\) is the value of \(\lambda\) at \(\theta=0\). Find net electric dipole moment for this charge distribution.

Short Answer

Expert verified
The net electric dipole moment is \( \pi R^2 \lambda_0 \) along the x-axis.

Step by step solution

01

Understanding the Problem

We need to find the net electric dipole moment for a ring with a specified linear charge density. The given linear charge density is a function of the angle \( \theta \), specifically \( \lambda = \lambda_{0} \cos \theta \). The radius of the ring is \( R \). Electric dipole moments involve the separation of positive and negative charges and are typically calculated as an integral when dealing with a continuous charge distribution on a geometric shape.
02

Formula for Electric Dipole Moment

The electric dipole moment \( \mathbf{p} \) of a continuous charge distribution is given by the integral \( \mathbf{p} = \int \, r \, dQ \), where \( r \) is the position vector of the charge element \( dQ \) relative to the reference point and extends over the entire charge distribution. In this case, since the distribution is circular, we express \( dQ \) in terms of \( \lambda \).
03

Express \( dQ \) in Terms of \( \theta \)

The charge element \( dQ \) can be expressed as \( dQ = \lambda \, R \, d\theta \). Substituting the given \( \lambda \), we get \( dQ = \lambda_{0} \cos \theta \times R \times d\theta \).
04

Determine the Position Vector

For a point on the ring parameterized by angle \( \theta \), the position vector is \( \mathbf{r} = (R \cos \theta, R \sin \theta) \). When calculating the dipole moment, we typically want to use the \( x \) and \( y \) components separately, denoted as \( p_x \) and \( p_y \).
05

Calculate \( p_x \)

The \( x \)-component of the dipole moment is \( p_x = \int R \cos \theta \, dQ = R^2 \lambda_0 \int_{0}^{2\pi} \cos^2 \theta \, d\theta \). Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \), this evaluates to \( R^2 \lambda_0 \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_{0}^{2\pi} = \pi R^2 \lambda_0 \).
06

Calculate \( p_y \)

The \( y \)-component of the dipole moment is \( p_y = \int R \sin \theta \, dQ = R^2 \lambda_0 \int_{0}^{2\pi} \cos \theta \, \sin \theta \, d\theta \). This integral evaluates to zero because it is symmetric around the \( x \)-axis, and the contributions from \( 0 \) to \( \pi \) and \( \pi \) to \( 2\pi \) cancel each other out.
07

Find the Net Dipole Moment

The net dipole moment \( \mathbf{p} \) of the ring is a vector with components \( p_x = \pi R^2 \lambda_0 \) and \( p_y = 0 \). This results in the dipole moment vector \( \mathbf{p} = (\pi R^2 \lambda_0, 0) \), indicating it lies entirely along the \( x \)-axis.

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Key Concepts

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

Linear Charge Density
Linear charge density is a measure of how much electric charge is distributed along a line, which in this case is the circumference of a circle. For a thin ring with radius \( R \), the linear charge density \( \lambda \) is specified by the function \( \lambda = \lambda_{0} \cos \theta \). Here, \( \lambda_{0} \) represents the maximum charge density at the position where the angle \( \theta \) is zero.
Notice how \( \lambda \) varies with \( \theta \); it's a function that changes from positive to negative, corresponding to the geometrical shape's orientation.
  • \( \lambda \) can be seen as how concentrated the charge is at different angles.
  • At \( \theta = 0 \), the charge density is at its peak as \( \cos \theta = 1 \).
  • At \( \theta = \frac{\pi}{2} \), the charge density drops to zero since \( \cos \frac{\pi}{2} = 0 \).
  • This cyclic nature continues around the ring.
Understanding linear charge density helps in visualizing how charges are spread out over the ring.
Circular Charge Distribution
A circular charge distribution refers to how charge is distributed along the curve of a circle. In this exercise, a non-conducting ring is used, and charges are distributed according to a sinusoidal function \( \lambda = \lambda_{0} \cos \theta \). This function implies that the charge distribution is not uniform around the ring; instead, it peaks and dips as one moves around the circle.
  • This specific form of distribution creates what is known as a dipole along certain axes like the x-axis here, as the net charge is zero but still results in a dipole moment.
  • The symmetry of the circular shape simplifies calculations considerably, especially if one exploits properties like even-odd functions while integrating.
  • By understanding this distribution, one can predict how the resulting electric fields will behave at large distances from the ring, further working into calculations such as python moments.
The ring's axial symmetry plays a crucial role in simplifying our physical understanding and mathematical treatment of this problem.
Integral Calculation
Integrating over a circular charge distribution requires translating continuous distributions into manageable mathematical expressions. In this problem, we calculate the electric dipole moment using integrals over the continuous charge distribution on the ring.
To find the dipole moment, we employ the formula:\[ \mathbf{p} = \int \mathbf{r} dQ \] where \( \mathbf{r} \) is the position vector and \( dQ \) the infinitesimal charge element on the ring.
  • The integral considers rotation around the angle \( \theta \), simplifying to evaluating integrals of trigonometric functions over a full period \( 0 \leq \theta \leq 2\pi \).
  • In this exercise, calculating \( p_x \) involves integrating \( \cos^2 \theta \), which requires using a trigonometric identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \).
  • Similarly, \( p_y \) relies on \( \sin \theta \cos \theta \), which when integrated over a symmetric interval cancels to zero due to its even-odd nature around the axis.
This comprehensive understanding of integrals, particularly with trigonometric identities, allows us to evaluate complex distributions effectively.
Position Vector Decomposition
The position vector is a crucial element when computing the electric dipole moment of a charge distribution. In this scenario, for a ring parameterized by an angle \( \theta \), the position vector \( \mathbf{r} \) is expressed in two dimensions as:\[ \mathbf{r} = (R \cos \theta, R \sin \theta) \]This decomposition into \( x \) and \( y \) components is key since calculations of the dipole moment consider each direction separately.
  • Each component contributes differently to the net dipole moment, revealing asymmetries or cancellations in the charge distribution.
  • The horizontal component \( R \cos \theta \) plays a significant role in the net dipole as we seek \( p_x \) whereas the vertical \( R \sin \theta \) contributes to \( p_y \).
  • Understanding this decomposition helps in predicting component-wise influences and makes the exertion of force calculable in vector fields.
By breaking down the position vector, we leverage symmetry and simplify calculating the components of the dipole moment.

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Most popular questions from this chapter

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of \(2.50 \mathrm{~mm}\) away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration-time and velocity-time graphs of the released proton's motion.

Torque on a Dipole. An electric dipole with dipole moment \(\vec{p}\) is in a uniform electric field \(\overrightarrow{\boldsymbol{E}}\). (a) Find the orientations of the dipole for which the torque on the dipole is zero. (b) Which of the orientations in part (a) is stable, and which is unstable? (Hint: Consider a small displacement away from the equilibrium position and see what happens.) (c) Show that for the stable orientation in part (b), the dipole's own electric field tends to oppose the external field.

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 \(\alpha=30.0^{\circ}\)

(a) What must the charge (sign and magnitude) of a \(1.3 \mathrm{~g}\) particle be for it to remain stationary when placed in a downward-directed electric field of magnitude \(650 \mathrm{~N} / \mathrm{C}\) ? (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, travelling \(4.50 \mathrm{~m}\) in the first \(3.00 \mu s\) after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.
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