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Chapter 21: Problem 78

Three point charges are placed on the \(y\) -axis: a charge \(q\) at \(y=a,\) a charge \(-2 q\) at the origin, and a charge \(q\) at \(y=-a\) . Such an arrangement is called an electric quadrupole. (a) Find the magnitude and direction of the electric field at points on the positive \(x\) -axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for \(x \gg a\) . Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole.

Short Answer

Expert verified
The electric field on the x-axis for large x is \( E \approx \frac{-3kqa^2}{x^4} \), decaying faster than a point charge or dipole field.

Step by step solution

01

Identify the Problem

We need to find the electric field generated by three point charges along the positive x-axis. These charges form an electric quadrupole with charges at different points on the y-axis.
02

Understand the Charge Arrangement

We have three point charges: +q at y = +a, -2q at y = 0, and +q at y = -a. We are interested in the electric field on the positive x-axis.
03

Electric Field Due to a Point Charge

The electric field due to a point charge q at a distance r is given by \[ E = \frac{k q}{r^2} \] where k is Coulomb's constant (8.99 脳 10^9 Nm虏/C虏) and r is the distance from the charge.
04

Calculate Distance to Each Charge

For a point (x,0) on the x-axis, the distances to the charges (x,卤a) and (x,0) are: - From +q at (0,a): \[ r_1 = \sqrt{x^2 + a^2} \] - From -2q at (0,0): \[ r_2 = x \] - From +q at (0,-a): \[ r_3 = \sqrt{x^2 + a^2} \]
05

Calculate Electric Fields from Each Charge

The magnitude of electric fields at (x,0) due to each charge is: - From +q at (0,a): \[ E_1 = \frac{k q}{r_1^2} \] - From -2q at (0,0): \[ E_2 = \frac{-2k q}{x^2} \] - From +q at (0,-a): \[ E_3 = \frac{k q}{r_3^2} \]
06

Find Net Electric Field

Because we are on the x-axis, the vertical components of the electric field due to the charges at (0,a) and (0,-a) will cancel out. The net electric field on the x-axis is: \[ E = E_1 + E_3 + E_2 \] \[ E = \frac{k q}{\sqrt{x^2 + a^2}^2} + \frac{k q}{\sqrt{x^2 + a^2}^2} - \frac{2 k q}{x^2} \]
07

Simplify for Large x (Binomial Approximation)

For \(x \gg a\), use the approximation \( (1 + \frac{a^2}{x^2})^{-3/2} \approx 1 - \frac{3a^2}{2x^2} \):\[ E \approx \frac{2kq}{x^2} \left(1 - \frac{3a^2}{2x^2}\right) - \frac{2kq}{x^2} \]\[ E \approx \frac{-3kqa^2}{x^4} \]This is much weaker than the \(1/x^2\) dependence of a single charge and highlights the weak field for distant points characteristic of quadrupoles.
08

Compare with Point Charge and Dipole

The electric field of a quadrupole decays as \( \frac{1}{x^4} \), which is faster than a point charge鈥檚 \( \frac{1}{x^2} \) decay and a dipole鈥檚 \( \frac{1}{x^3} \) decay.

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Key Concepts

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

Binomial Expansion
The binomial expansion is a powerful tool in mathematics used to approximate expressions that involve powers of binomials. Using this technique can simplify calculations, especially when dealing with powers raised to high exponents.

In the context of an electric quadrupole, the binomial expansion comes into play when simplifying the expression for the electric field at a point far away from the charges. Typically, we approximate terms like \((1 + \frac{a^2}{x^2})^{-3/2}\) using the binomial expansion formula:
  • \((1 + y)^n \approx 1 + ny + \frac{n(n-1)}{2!}y^2 + \ldots\)
For our scenario, when \(x \gg a\), this simplifies to just \(1 - \frac{3a^2}{2x^2}\), making our calculations manageably concise.

This approach highlights how an electric quadrupole's field strength diminishes faster than simpler charge arrangements.
Electric Field
An electric field is a vector field that surrounds electric charges and exerts a force on other charges within the field. The strength and direction of the electric field created by a point charge can be calculated using the formula:
  • \(E = \frac{kq}{r^2}\)
Here, \(E\) represents the electric field magnitude, \(k\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge to the point of interest.

The direction of the electric field is always directed away from positive charges and toward negative charges. When dealing with multiple charges, you must consider the vector sum of each individual field to find the net electric field at a particular point. For an electric quadrupole, the field becomes markedly weaker as you move further away, diminishing more rapidly than an isolated charge or dipole due to the arrangement's symmetry.
Coulomb's Constant
Coulomb's constant, denoted as \(k\), is a fundamental value in electrostatics aiding in measuring the force between two charges. It appears in the equation for calculating the electric field as well:
  • \( k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)
This constant stems from Coulomb's law, which states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In electric quadrupole problems, Coulomb's constant facilitates calculating individual forces and fields created by each charge. This constant is crucial for precise, accurate computations of electric interactions, allowing us to predict electric field behavior even in complex configurations.
Electric Dipole
An electric dipole consists of two equal and opposite charges separated by a small distance. It has a significant impact on electric field behavior, especially when studying charge distributions like a quadrupole.

The electric field of a dipole diminishes more quickly with distance than a single point charge and can be expressed as:
  • \(E \propto \frac{1}{x^3}\)
This rapid decay is due to the opposite charges effectively neutralizing each other at a distance.

Unlike a dipole, an electric quadrupole contains four charges arranged symmetrically, leading to an even faster decay of its electric field at large distances, reducing as \(\frac{1}{x^4}\). This stark contrast showcases how charge configuration affects field strength and decay, crucial for understanding electric field interactions in different setups.

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Most popular questions from this chapter

CP Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius \(r=5.29 \times 10^{-11} \mathrm{m}\) around a stationary proton. What is the speed of the electron in its orbit?

CP Two small spheres with mass \(m=15.0\) g are hung by silk threads of length \(L=1.20 \mathrm{m}\) from a common point (Fig. P21.68). When the spheres are given equal quantities of negative charge, so that \(q_{1}=q_{2}=q,\) each thread hangs at \(\theta=25.0^{\circ}\) from the vertical. (a) Draw a diagram showing the forces on each sphere. Treat the spheres as point charges. (b) Find the magnitude of \(q .\) (c) Both threads are now shortened to length \(L=0.600 \mathrm{m},\) while the charges \(q_{1}\) and \(q_{2}\) remain unchanged. What new angle will each thread make with the vertical? (Hint: This part of the problem can be solved numerically by using trial values for \(\theta\) and adjusting the values of \(\theta\) until a self- consistent answer is obtained.)

BIO Electrophoresis. Electrophoresis is a process used by biologists to separate different biological molecules (such as proteins) from each other according to their ratio of charge to size. The materials to be separated are in a viscous solution that produces a drag force \(F_{\mathrm{D}}\) proportional to the size and speed of the molecule. We can express this relation- ship as \(F_{\mathrm{D}}=K R v,\) where \(R\) is the radius of the molecule (modeled as being spherical), \(v\) is its its speed, and \(K\) is a constant that depends on the viscosity of the solution. The solution is placed in an external electric field \(E\) so that the electric force on a particle of charge \(q\) is \(F=q E\) . (a) Show that when the electric field is adjusted so that the two forces (electric and viscous drag) just balance, the ratio of \(q\) to \(R\) is \(K v / E\) . (b) Show that if we leave the electric field on for a time \(T,\) the distance \(x\) that the molecule moves during that time is \(x=(E T / k)(q / R)\) . (c) Suppose you have a sample containing three different biological molecules for which the molecular ratio \(q / R\) for material 2 is twice that of material 1 and the ratio for material 3 is three times that of material 1. Show that the distances migrated by these molecules after the same amount of time are \(x_{2}=2 x_{1}\) and \(x_{3}=3 x_{1}\) . In other words, material 2 travels twice as far as material \(1,\) and material 3 travels three times as far as material \(1 .\) Therefore, we have separated these molecules according to their ratio of charge to size. In practice, this process can be carried out in a special gel or paper, along which the biological molecules migrate. (Fig. P21.94). The process can be rather slow, requiring several hours for separations of just a centimeter or so.

CP Two identical spheres are each attached to silk threads of length \(L=0.500 \mathrm{m}\) and hung from a common point (Fig. P21.68). Each sphere has mass \(m=8.00 \mathrm{g} .\) The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_{1},\) and the other a different positive charge \(q_{2} ;\) this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta=20.0^{\circ}\) with the vertical. (a) Draw a free-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the information you have been given, what can you say about the magnitudes of \(q_{1}\) and \(q_{2} ?\) Explain your answers. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of \(30.0^{\circ}\) with the vertical. Determine the original charges. (Hint: The total charge on the pair of spheres is conserved.)

CP An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 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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