/*! 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 97 CALC Negative charge \(-Q\) is d... [FREE SOLUTION] | 91Ó°ÊÓ

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CALC 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 the x and y components of the electric field are \(\frac{2k_e Q}{\pi a^2}\), directed in the negative directions.

Step by step solution

01

Understanding the Setup

We have a quarter-circle in the first quadrant, spanning from the positive x-axis to the positive y-axis, with its center of curvature at the origin. The quarter-circle has a radius of \(a\) and carries a total negative charge \(-Q\). We need to find the electric field components at the origin.
02

Infinitesimal Charge Element

Consider a small charge element \(dq\) on the quarter-circle. Since charge \(-Q\) is distributed uniformly, the linear charge density \(\lambda\) is \(\lambda = \frac{-Q}{\frac{\pi a}{2}} = \frac{-2Q}{\pi a}\). The charge element \(dq\) over an infinitesimal arc length \(ds = a d\theta\) is \(dq = \lambda \, ds = \frac{-2Q}{\pi a} \, a d\theta\).
03

Determine Electric Field Contribution

The contribution to the electric field \(d\mathbf{E}\) at the origin from \(dq\) is given by Coulomb's law: \(dE = \frac{k_e \, |dq|}{a^2}\), where \(k_e\) is Coulomb's constant. The direction of \(d\mathbf{E}\) at the origin from \(dq\) at angle \(\theta\) is radially inward, forming an angle \(\theta\) with the negative x-axis.
04

Components of the Electric Field

Decompose \(d\mathbf{E}\) into its components: \(dE_x = dE \cos\theta\) and \(dE_y = dE \sin\theta\). Substitute \(dE = \frac{k_e \, 2Q \, d\theta}{\pi a}\) to get the components: \(dE_x = \frac{k_e \, 2Q}{\pi a^2} \, \cos\theta \, d\theta\) and \(dE_y = \frac{k_e \, 2Q}{\pi a^2} \, \sin\theta \, d\theta\).
05

Integrate to Find Total Electric Field Components

Integrate \(dE_x\) from \(0\) to \(\frac{\pi}{2}\): \[E_x = \int_0^{\pi/2} \frac{k_e \, 2Q}{\pi a^2} \, \cos\theta \, d\theta = \frac{k_e \, 2Q}{\pi a^2} \left[ \sin\theta \right]_0^{\pi/2} = \frac{k_e \, 2Q}{\pi a^2} \cdot 1 = \frac{2k_e Q}{\pi a^2}\]Integrate \(dE_y\) from \(0\) to \(\frac{\pi}{2}\):\[E_y = \int_0^{\pi/2} \frac{k_e \, 2Q}{\pi a^2} \, \sin\theta \, d\theta = \frac{k_e \, 2Q}{\pi a^2} \left[ -\cos\theta \right]_0^{\pi/2} = \frac{k_e \, 2Q}{\pi a^2} \cdot 1 = \frac{2k_e Q}{\pi a^2}\]
06

Conclusion

The x-component of the net electric field at the origin is \(E_x = \frac{2k_e Q}{\pi a^2}\) directed in the negative x-direction, and the y-component is \(E_y = \frac{2k_e Q}{\pi a^2}\) directed in the negative y-direction in accordance with the field direction inward towards the charge distribution.

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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 foundational in understanding electric fields. It describes how electric charges interact at a distance. Imagine electric charge as a tiny invisible thread tying two charged spheres, either pulling them together or pushing them apart. This interaction is what we calculate through Coulomb's Law.
For point charges, the force between them is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:\[F = k_e \frac{|q_1 q_2|}{r^2}\]where:
  • \(F\) is the magnitude of the force between the charges
  • \(k_e\) is Coulomb's constant \(\approx 8.99 \times 10^9 \text{Nm}^2/\text{C}^2\)
  • \(q_1\) and \(q_2\) are the amounts of the charges
  • \(r\) is the distance between the charges
In our problem, we have a distribution of charge along a quarter-circle rather than point charges. We apply Coulomb's Law to calculate the effect of each infinitesimally small piece of charge along the curve. This requires summing, or rather integrating, these small effects to find the total electric field at a specific point.
Charge Distribution
Charge distribution refers to how electric charges are spread across a given area or line, influencing the surrounding electric field. In uniform charge distribution, like the one in our quarter-circle problem, each small segment of the arc contributes equally to the electric field at a point.
The linear charge density \(\lambda\) describes how much charge resides along a line segment, calculated as the total charge \(-Q\) divided by the length of the distribution. For our quarter-circle:\[\lambda = \frac{-Q}{\frac{\pi a}{2}} = \frac{-2Q}{\pi a}\]This formula helps us define a small charge element \(dq\) at a point \(\theta\) on the arc:\[dq = \lambda \cdot ds = \frac{-2Q}{\pi a} \cdot a d\theta\]By understanding how charge is distributed, we can determine how it influences the electric field. In essence, the shape, size, and nature of the distribution provides the framework for tackling complex problems involving electric fields.
Vector Integration
Vector integration allows us to calculate total effects from distributed sources in multiple dimensions. In our scenario, small charge segments on the quarter-circle each create electric fields pointing towards the origin. We need to add these contributions to determine the total electric field.
The magnitude of the electric field contributed by each segment is determined using Coulomb's law. Its direction points radially inward, aligning with the radius of the circle. Thus, we break the little electric field vectors into components along the x and y axes:\[dE_x = dE \cos\theta, \quad dE_y = dE \sin\theta\]By integrating these components over all infinitesimal charge elements along the quarter-circle, from angle \(0\) to \(\pi/2\), we find the total electric field. The integrations are:\[E_x = \int_0^{\pi/2} \frac{k_e 2Q}{\pi a^2} \cos\theta \, d\theta = \frac{2k_e Q}{\pi a^2}\]\[E_y = \int_0^{\pi/2} \frac{k_e 2Q}{\pi a^2} \sin\theta \, d\theta = \frac{2k_e Q}{\pi a^2}\]Vector integration transforms a complex, distributed charge setup into manageable calculations, yielding the resultant electric field at a focal point.

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Most popular questions from this chapter

A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x y\) -coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\) -axis at \(x=4.00 \mathrm{cm} .\) (a) If a third charge \(q_{3}=\) \(+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{cm}, y=3.00 \mathrm{cm}\) find the \(x\) - and \(y\) -components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

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

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(r\) in the direction of (a) the field point at \(x=0, \quad y=-1.35 \mathrm{m}\) ; (b) the field point at \(x=12.0 \mathrm{cm}, y=12.0 \mathrm{cm} ;(\mathrm{c})\) the field point at \(x=-1.10 \mathrm{m},\) \(y=2.60 \mathrm{m} ?\) Express your results in terms of the unit vectors \(\hat{\boldsymbol{\imath}}\) and \(\hat{\boldsymbol{J.}}\)

Particles in a Gold Ring. You have a pure (24 karat) gold ring with mass 17.7 \(\mathrm{g}\) . Gold has atomic mass of 197 \(\mathrm{g} / \mathrm{mol}\) and an atomic number of \(79 .\) (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?

Electric Field of the Earth. The earth has a net electric charge that causes a field at points near its surface equal to 150 \(\mathrm{N} / \mathrm{C}\) and directed in toward the center of the earth. (a) What magnitude and sign of charge would a \(60-\mathrm{kg}\) 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 \(\mathrm{m} ?\) Is use of the earth's electric field a feasible means of flight? Why or why not?

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