/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 Negative charge \(-Q\) is distri... [FREE SOLUTION] | 91Ó°ÊÓ

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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
Both components are \( E_x = E_y = \frac{k_e 2Q}{\pi a^2} \) towards the origin.

Step by step solution

01

Understanding the Problem

We have a quarter-circle of radius \( a \) in the first quadrant centered at the origin, carrying a uniform negative charge \( -Q \). We are to find the \( x \)- and \( y \)-components of the electric field at the origin due to this charge distribution.
02

Using Symmetry

Given the symmetry of the problem, the electric field at the origin should have components along both the \( x \) and \( y \) axes. Due to the symmetry of the charge distribution along the quarter-circle in the first quadrant, contributions to the horizontal (\( x \)) and vertical (\( y \)) components will be symmetric.
03

Electric Field Element Contribution

Consider a small charge element \( dq \) on the arc, creating an electric field \( d\mathbf{E} \) at the origin. The infinitesimal charge element can be expressed as \( dq = \frac{-Q}{\frac{\pi}{2} a} ad\theta = \frac{-2Q}{\pi} d\theta \), where \( \theta \) is the angle made with the positive x-axis.
04

Expression for Electric Field

The magnitude of the electric field produced by a charge element \( dq \) at distance \( a \) is given by \( dE = \frac{k_e |dq|}{a^2} \), where \( k_e \) is Coulomb's constant and \( dE \) is along the radius of the circle. Since \( dE = \frac{k_e 2Q}{\pi a^2} d\theta \), it has both \( x \) and \( y \) components due to its circular symmetry.
05

Determine Components of dE

The \( x \)-component of \( dE \) is \( dE_x = dE \cos \theta = \frac{k_e 2Q}{\pi a^2} \cos \theta \, d\theta \) and the \( y \)-component is \( dE_y = dE \sin \theta = \frac{k_e 2Q}{\pi a^2} \sin \theta \, d\theta \).
06

Integration to Find Net x and y Components

Integrate \( dE_x \) and \( dE_y \) from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \):- For \( E_x \): \[ E_x = \int_{0}^{\pi/2} \frac{k_e 2Q}{\pi a^2} \cos \theta \, d\theta = \frac{k_e 2Q}{\pi a^2} \left[ \sin \theta \right]_0^{\pi/2} = \frac{k_e 2Q}{\pi a^2} (1 - 0) = \frac{k_e 2Q}{\pi a^2} \]- For \( E_y \):\[ E_y = \int_{0}^{\pi/2} \frac{k_e 2Q}{\pi a^2} \sin \theta \, d\theta = \frac{k_e 2Q}{\pi a^2} \left[ -\cos \theta \right]_0^{\pi/2} = \frac{k_e 2Q}{\pi a^2} (0 + 1) = \frac{k_e 2Q}{\pi a^2} \]
07

Conclusion

The net electric field components due to the quarter-circle at the origin are \( E_x = \frac{k_e 2Q}{\pi a^2} \) and \( E_y = \frac{k_e 2Q}{\pi a^2} \). Due to the negative charge, these components are directed towards negative x and y directions, but we express magnitudes here.

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Key Concepts

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

Coulomb's Law
Coulomb's law is a fundamental principle that describes the force between two point charges. This law states that the magnitude of the electric force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This can be mathematically expressed as:\[ F = \frac{k_e |q_1 q_2|}{r^2} \]where
  • F is the magnitude of the force between the charges,
  • ke is Coulomb's constant \(\sim 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\),
  • |q1| and |q2| are the magnitudes of the charges, and
  • r is the distance between the centers of the two charges.
In this particular exercise, we utilize Coulomb's law to determine the electric field created by small elements of charge along a quarter-circle.
Instead of focusing on force, our goal is to find the electric field at the origin, which requires evaluating the effect of each differential charge element using Coulomb's law.
Symmetry in Physics
Symmetry plays a crucial role in simplifying complex physics problems, particularly in calculating electric fields. Symmetry can often reveal inherent properties that cut down computational complexity. In this exercise, the charge distribution is symmetric about both
  • the x-axis
  • the y-axis
The quarter-circle of negative charge presents such a symmetry in the first quadrant.
This symmetry allows us to infer that the electric field contributions from various charge elements will have components both along the x and y axes. Since the arrangement is uniform, integrals over symmetric regions lead to equivalent results for components in directions of symmetry.
In this way, symmetry helps in simplifying the math required to find separate electric field components.
Integral Calculus
Integral calculus is a branch of mathematics that deals with the accumulation of quantities, such as areas under curves or, in this case, the total electric field. Integral calculus allows for the summation of infinitesimally small contributions of electric field due to a
  • continuous charge distribution
This method is essential when evaluating fields where charge is not isolated, but spread over a region. In our exercise, each small element of charge along the quarter-circle contributes a slice of the electric field at the origin.
By integrating these small contributions around the angle from 0 to \(\frac{\pi}{2}\), we calculate the total electric field components in the x and y directions. Calculating these integrals tells us how fields from all charge elements add up, demonstrating how continuous mathematical ideas consolidate to quantifiable physical quantities.
Charge Distribution
Charge distribution describes how electric charge is spread across a region of space. The distribution can be uniform, which means distributed evenly, or non-uniform, where charge density varies across space. In our scenario, the distribution is uniform around a quarter-circle, meaning the charge
  • is spread evenly along this arc in the first quadrant,
  • maintaining constant density,
  • denoted by the variable \(-Q\).
This constant density facilitates calculations since each specific element of charge on the arc, denoted by dq, is evenly contributing to the net electric field at the origin.
Understanding charge distribution is vital to identifying how each contributing element affects the entire system, thereby enabling the calculation of resultant fields and forces.

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Most popular questions from this chapter

A nerve signal is transmitted through a neuron when an excess of \(Na^+\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately 10.0 \(\mu\)m in diameter, and measurements show that about 5.6 \(\times \space 10^{11} \space Na^+\)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-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-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 cm below the skin? (c) Certain sharks can respond to electric fields as weak as 1.0 \(\mu N/C\). How far from this segment of axon could a shark be and still detect its electric field?

(a) An electron is moving east in a uniform electric field of 1.50 N\(/\)C directed to the west. At point \(A\), the velocity of the electron is 4.50 \(\times 10^5\) m\(/\)s toward the east. What is the speed of the electron when it reaches point B, 0.375 m east of point A? (b) A proton is moving in the uniform electric field of part (a). At point \(A\), the velocity of the proton is 1.90 \(\times 10^4\) m\(/\)s, east. What is the speed of the proton at point \(B\)?

A very long, straight wire has charge per unit length \(3.20 \times 10^{-10}\) C/m. At what distance from the wire is the electric-field magnitude equal to 2.50 N\(/\)C?

Excess electrons are placed on a small lead sphere with mass 8.00 g so that its net charge is -3.20 x 10\(^{-9}\)C. (a) Find the number of excess electrons on the sphere. (b) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol.

The earth has a net electric charge that causes a field at points near its surface equal to 150 N\(/\)C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth's electric field a feasible means of flight? Why or why not?

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