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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

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.

A semicircle of radius \(a\) is in the first and second quadrants, with the center of curvature at the origin. Positive charge \(+Q\) is distributed uniformly around the left half of the semicircle, and negative charge \(-Q\) is distributed uniformly around the right half of the semicircle (Fig. P21.84). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge?

(a) What must the charge (sign and magnitude) of a \(1.45 \mathrm{~g}\) particle be for it to remain stationary when placed in a downwarddirected 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?

A nerve signal is transmitted through a neuron when an excess of \(\mathrm{Na}^{+}\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately \(10.0 \mu \mathrm{m}\) in diameter, and measurements show that about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions per meter (each of charge \(+e\) ) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of point charges moving along the axon. Consider a \(0.10 \mathrm{~mm}\) length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a \(0.10 \mathrm{~mm}\) length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is \(5.00 \mathrm{~cm}\) below the skin? (c) Certain shark can respond to electric fields as weak as \(1.0 \mu \mathrm{N} / \mathrm{C}\). How far from this segment of axon could a shark be and still detect its electric field?

An electric dipole with dipole moment \(\vec{p}\) is in a uniform external electric field \(E\). (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 rotation 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.
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