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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 at points on the x-axis is directed in the negative x-direction with magnitude approximated by \( E \approx -\frac{4kqa^2}{x^4} \) for \( x \gg a \).

Step by step solution

01

Setting up the problem

We have three point charges on the y-axis: a charge \( q \) at \( y = a \), a charge \( -2q \) at the origin (\( y = 0 \)), and a charge \( q \) at \( y = -a \). We are tasked with finding the electric field on the positive x-axis due to this electric quadrupole arrangement. We will consider a point \( P \) on the x-axis located at \( (x, 0) \).
02

Expression for Electric Field from a Point Charge

The electric field \( \mathbf{E} \) due to a point charge \( q \) at a distance \( r \) is given by \( \mathbf{E} = \frac{kq}{r^2} \hat{r} \), where \( k \) is Coulomb's constant and \( \hat{r} \) is the unit vector from the charge to the point. This formula will be used to determine the contribution of each individual charge to the field at point \( P \).
03

Calculate Contributions of Each Charge

Calculate the distance from each charge to the point \( P \):- For the charge \( q \) at \( y = a \): Distance = \( \sqrt{x^2 + a^2} \)- For the charge \( -2q \) at the origin: Distance = \( x \)- For the charge \( q \) at \( y = -a \): Distance = \( \sqrt{x^2 + a^2} \).
04

Electric Field Due to Each Charge at Point P

For the charge at \( y = a \), the field \( \mathbf{E}_1 = \frac{kq}{(x^2 + a^2)} \left(-\frac{x}{\sqrt{x^2 + a^2}}, \frac{a}{\sqrt{x^2 + a^2}}\right) \).For the charge at \( y = 0 \), the field \( \mathbf{E}_2 = \frac{-2kq}{x^2} (1, 0) \).For the charge at \( y = -a \), the field \( \mathbf{E}_3 = \frac{kq}{(x^2 + a^2)} \left(-\frac{x}{\sqrt{x^2 + a^2}}, -\frac{a}{\sqrt{x^2 + a^2}}\right) \).
05

Calculation of Net Electric Field on the x-axis

Sum the x and y components of the electric fields from all three charges at \( P \):- Total x-component: \[ E_{x} = \frac{kq(-2x)}{(x^2 + a^2)^{3/2}} + \frac{-2kq}{x^2} \]- Total y-component cancels out due to symmetry.Therefore, the net electric field \( \mathbf{E} = -\frac{2kqx}{(x^2 + a^2)^{3/2}} - \frac{2kq}{x^2}. \) Then use the negative x-direction direction for field.
06

Binomial Approximation for x >> a

Using binomial expansion, for \( x \gg a \), the approximation \( (x^2 + a^2)^{3/2} \approx x^3 \). Plugging this into the previous expression gives \( E_x \approx -\frac{4kqa^2}{x^4} \), where \( k = \frac{1}{4\pi\varepsilon_0} \).
07

Comparing with Point Charge and Dipole

For a point charge, the electric field magnitude decreases as \( \frac{1}{x^2} \), and for a dipole, the electric field magnitude decreases as \( \frac{1}{x^3} \). In contrast, the electric field of a quadrupole decreases as \( \frac{1}{x^4} \), which indicates a rapid decrease in electric field strength with distance compared to a point charge or a dipole.

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Key Concepts

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

Electric Field
The electric field is an essential concept in understanding how charges interact in space. Imagine it as an invisible force field around a charge that exerts a push or a pull on other charges. This field is generated by any charge, positive or negative, and its direction is determined by the type of charge:
  • For a positive charge, the electric field points away from the charge.
  • For a negative charge, it points toward the charge.
The strength of the electric field, represented as \( E \), is influenced by the amount of charge and the distance from it. According to the formula from Coulomb's law, for a single point charge \( q \), the electric field at a distance \( r \) is given by \( E = \frac{kq}{r^2} \), where \( k \) is Coulomb’s constant. In the specific scenario of an electric quadrupole on the y-axis, multiple charges create a more complex field. Each charge contributes to the net field at a point, and their distances and configuration affect the field's total strength and direction.
Binomial Expansion
Binomial expansion is a mathematical tool used for simplifying expressions. It's much like unfolding a complex combination into simpler parts. When dealing with large values of distance compared to another magnitude in a formula, like \( x \gg a \), binomial expansion becomes very useful. For any binomial expression \((1 + b)^n\), it approximates as \(1 + nb + \frac{n(n-1)}{2!}b^2\) for small values of \( b \).
  • Helps in approximating complex equations where one variable is much larger than the others.
  • The expansion's primary use is to make complicated expressions more approachable.
In the context of our electric quadrupole exercise, the binomial expansion aids in deriving an approximate expression for the electric field when the point of interest on the x-axis is far away from the quadrupole system. For instance, \((x^2 + a^2)^{3/2}\) simplifies to \(x^3\) using binomial expansion, greatly simplifying the calculation of the electric field strength at distant points.
Coulomb's Law
Coulomb's law is the foundation upon which electric field calculations are based. It provides a simple yet profound insight into how charged particles interact. The law states that the force \( F \) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance \( r \) between them:\[ F = \frac{k |q_1 q_2|}{r^2} \]Where:
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance separating the charges.
  • \( k \) is the Coulomb's constant \( \left( \approx 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \right) \).
This law not only predicts the strength of the force but also its direction, which is along the line joining the two charges. In our exercise's scenario, Coulomb's law helps in calculating the individual contributions to the electric field from each charge in the quadrupole. By understanding this, one can better comprehend how these fields interact and add up to form the overall electric field in the system, especially as evaluated at a specific point on the x-axis.

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Most popular questions from this chapter

Two thin rods of length \(L\) lie along the \(x\)-axis, one between \(x=a / 2\) and \(x=a / 2+L\) and the other between \(x=-a / 2\) and \(x=-a / 2-L .\) Each rod has positive charge \(Q\) distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive \(x\)-axis. (b) Show that the magnitude of the force that one rod exerts on the other is $$ F=\frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2 L)}\right] $$ (c) Show that if \(a \gg L, \quad\) the magnitude of this force reduces to \(F=Q^{2} / 4 \pi \epsilon_{0} a^{2} . \quad\) (Hint: Use the expansion \(\ln (1+z)=z-\) \(z^{2} / 2+z^{3} / 3-\cdots\), valid for \(|z| \ll 1\). Carry all expansions to at least order \(L^{2} / a^{2}\).) Interpret this result.

Four identical charges \(Q\) are placed at the corners of a square of side \(L\). (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges.

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, \(2.25 \mathrm{~cm}\) distant from the first, in a time interval of \(1.50 \times 10^{-6}\) s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

Three point charges are arranged on a line. Charge \(q_{3}=+5.00 \mathrm{nC}\) and is at the origin. Charge \(q_{2}=-3.00 \mathrm{nC}\) and is at \(x=+4.00 \mathrm{~cm}\). Charge \(q_{1}\) is at \(x=+2.00 \mathrm{~cm}\). What is \(q_{1}\) (magnitude and sign) if the net force on \(q_{3}\) is zero?

An average human weighs about \(650 \mathrm{~N}\). If two such generic humans each carried \(1.0\) coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their \(650-\mathrm{N}\) weight?

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