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Chapter 21: Problem 83

Negative charge \(-Q\) is distributed uniformly around a quarter-circle of radius \(a\) that lies in the first quadrant, with the center of curvature at the origin. Find the \(x\) - and \(y\) -components of the net electric field at the origin.

Short Answer

Expert verified
The x and y components of the net electric field at the origin, due to the uniformly distributed charge, are \(-Qk_e/(2a^2)\) and \(-Qk_e/(2a^2)\) respectively.

Step by step solution

01

Define the problem parameters

Firstly, let's denote the uniformly distributed charge as \( \lambda = -Q / (\pi a/2) \), as it is distributed around a quarter-circle with radius \(a\). The total angle covered by the quarter-circle is \(\pi/2\). The infinitesimal charge \(dq\) is then given by \( \lambda a d\theta \), where \(d\theta\) is the infinitesimal angle.
02

Set up the integral for the electric field contributions

Let's consider an infinitesimal charge located at an angle \(\theta\) from the positive x-axis. The electric field \(d\vec{E}\) due to this charge at the origin (which is also the center of curvature) is given by Coulomb's law, i.e., \( d\vec{E} = k_e \frac{dq}{a^2} \) where \( k_e \) is Coulomb's constant. The electric field has a magnitude of \(dE\) and direction along the line joining the infinitesimal charge and the origin. Using trigonometry, the x and y components can be expressed as \(dE_x = dE \sin\left(\theta\right)\) and \(dE_y = - dE \cos\left(\theta\right)\) respectively.
03

Evaluate the integral

To get the net electric field at the origin, we must integrate over the entire charge distribution. This results in: \(\vec{E}_x = \int_{0}^{\pi/2} dE_x\) and \(\vec{E}_y = \int_{0}^{\pi/2} dE_y\). After substituting \(d\vec{E}\) and evaluating the integral, we get \(\vec{E}_x = -Qk_e/(2a^2)\) and \(\vec{E}_y = -Qk_e/(2a^2)\).
04

Summarize the results

The net electric field at the origin due to the uniformly distributed negative charge on the quarter-circle is directed along the negative x and y axes (indicated by the negative signs), each having a magnitude of \(Qk_e/(2a^2)\).

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Key Concepts

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

Coulomb's Law
Understanding Coulomb's Law is essential when dealing with charges and electric fields. Named after Charles-Augustin de Coulomb, this fundamental principle of electromagnetism states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it is described by the formula:
\[ F = k_e \frac{|q_1 q_2|}{r^2} \]where
  • \(F\) is the magnitude of the electric force between the charges,
  • \(k_e\) is Coulomb's constant \((8.9875 \times 10^9 \text{ N m}^2/\text{C}^2)\),
  • \(q_1\) and \(q_2\) are the point charges,
  • \(r\) is the distance between the charges.
Coulomb's Law is crucial for determining the electric field produced by a charge distribution. In the context of the exercise, each infinitesimal charge segment \(dq\) contributes to the electric field at the origin. By integrating these contributions across the entire charge distribution, we can find the total electric field.
Uniform Charge Distribution
A uniform charge distribution implies that charge is spread evenly over a specified area, length, or volume. In the exercise, this distribution is along a quarter-circle, with the charge density \(\lambda\) being constant. It is defined by:
\[ \lambda = \frac{-Q}{\pi a / 2} \]where
  • \(Q\) is the total charge distributed over the length of the quarter-circle,
  • \(a\) is the radius of the quarter-circle,
  • \(\pi/2\) is the angle subtended by the quarter-circle in radians.
The concept of uniform distribution is significant because it simplifies the calculation of electric fields and forces. By knowing the charge density \(\lambda\), we can calculate the influence of infinitesimally small charges, \(dq = \lambda a d\theta\). This systematic approach allows us to determine how the entire configured charge affects a point in space, in this case, the origin.
Vector Components
In physics, breaking down vectors into components simplifies problems, especially when dealing with forces or electric fields. A vector in a 2D plane can be divided into two components鈥攁long the x-axis and y-axis.
In the exercise, the infinitesimal electric field \(d\vec{E}\), produced by each small charge \(dq\), is split into x and y components using trigonometry:
  • \(dE_x = dE \sin(\theta)\)
  • \(dE_y = - dE \cos(\theta)\)
Here, \(\theta\) is the angle from the x-axis.
The negative sign in \(dE_y\) indicates the direction of the field component opposing the standard positive y direction. Once decomposed, these components can be separately integrated over the interval corresponding to the quarter-circle \(0\text{ to }\pi/2\). This method reveals the net electric field at the origin, highlighting its direction and magnitude.

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Most popular questions from this chapter

A disk with radius \(R\) and uniform positive charge density \(\sigma\) lies horizontally on a tabletop. A small plastic sphere with mass \(M\) and positive charge \(Q\) hovers motionless above the center of the disk, suspended by the Coulomb repulsion due to the charged disk. (a) What is the magnitude of the net upward force on the sphere as a function of the height \(z\) above the disk? (b) At what height \(h\) does the sphere hover? Express your answer in terms of the dimensionless constant \(v \equiv 2 \epsilon_{0} M g /(Q \sigma) .\) (c) If \(M=100 \mathrm{~g}, Q=1 \mu \mathrm{C}, R=5 \mathrm{~cm},\) and \(\sigma=10 \mathrm{nC} / \mathrm{cm}^{2},\) what is \(h ?\)

A very long line of charge with charge per unit length \(+8.00 \mu \mathrm{C} / \mathrm{m}\) is on the \(x\) -axis and its midpoint is at \(x=0 .\) A second very long line of charge with charge per length \(-4.00 \mu \mathrm{C} / \mathrm{m}\) is parallel to the \(x\) -axis at \(y=10.0 \mathrm{~cm}\) and its midpoint is also at \(x=0 .\) At what point on the \(y\) -axis is the resultant electric field of the two lines of charge equal to zero?

A thin disk with a circular hole at its center, called an \(a n-\) nulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. \(\mathbf{P 2 1 . 8 7}\) ). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z\) plane, with its center at the origin. For an arbitrary point on the \(x\) -axis (the axis of the annulus), find the magnitude and direction of the electric field \(\overrightarrow{\boldsymbol{E}}\). Consider points both above and below the annulus. (c) Show that at points on the \(x\) -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\) -axis (but cannot move off the axis). The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of \(1.4 \times 10^{-8} \mathrm{~g}\). The drops will leave the nozzle and travel toward the paper at \(50 \mathrm{~m} / \mathrm{s}\) in a horizontal direction, passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops will then pass between parallel deflecting plates, \(2.0 \mathrm{~cm}\) long, where there is a uniform vertical electric field with magnitude \(8.0 \times 10^{4} \mathrm{~N} / \mathrm{C}\). Your team is working on the design of the charging unit that places the charge on the drops. (a) If a drop is to be deflected \(0.30 \mathrm{~mm}\) by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge? (b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of \(25 \mathrm{~m} / \mathrm{s},\) what \(q\) value is needed to achieve the same \(0.30 \mathrm{~mm}\) deflection?

Two thin rods, each with length \(L\) and total charge \(+Q,\) are parallel and separated by a distance \(a .\) The first rod has one end at the origin and its other end on the positive \(y\) -axis. The second rod has its lower end on the positive \(x\) -axis. (a) Explain why the \(y\) -component of the net force on the second rod vanishes. (b) Determine the \(x\) -component of the differential force \(d F_{2}\) exerted on a small portion of the second rod, with length \(d y_{2}\) and position \(y_{2},\) by the first rod. (This requires integrating over differential portions of the first rod, parameterized by \(\left.d y_{1} .\right)\) (c) Determine the net force \(\vec{F}_{2}\) on the second rod by integrating \(d F_{2 x}\) over the second rod. (d) Show that in the limit \(a \gg L\) the force determined in part (c) becomes \(\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2}}{a^{2}} \hat{\imath}\). (e) Determine the external work required to move the second rod from very far away to the position \(x=a\), provided the first rod is held fixed at \(x=0 .\) This describes the potential energy of the original configuration. (f) Suppose \(L=50.0 \mathrm{~cm}, a=10.0 \mathrm{~cm}, Q=10.0 \mu \mathrm{C},\) and \(m=500 \mathrm{~g}\). If the two rods are released from the original configuration, they will fly apart and ultimately achieve a particular relative speed. What is that relative speed?
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