/*! 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 70 A charge of \(-3.00 \mathrm{nC}\... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
Once calculated, the x and y components of the total force, as well as its magnitude and direction, will be presented. Note that the direction must account for the quadrant of the vector based on the directional components.

Step by step solution

01

Calculate the forces between charges

To start, we must calculate the forces between charges using Coulomb's law \(F = k \cdot \frac{{|q1 \cdot q2|}}{{r^2}}\), where \(k = 8.99 \times 10^9 N \cdot m^2/C^2\) is Coulomb's constant, \(q1\) and \(q2\) are the magnitudes of the charges and \(r\) is the distance between charges. The charge at the origin and the one at point (3cm, 4cm) forms a right triangle with the third charge, with a total distance of \(\sqrt{{(3cm)^2 + (4cm)^2}} = 5cm\). Compute \(F_{13} = k \cdot \frac{{|-3nC \cdot 5nC|}}{{(5cm)^2}}\) and \(F_{23} = k \cdot \frac{{|2nC \cdot 5nC|}}{{(4cm)^2}}\).
02

Find force components

Each force has an x-component and a y-component. For \(F_{13}\), since it's directed along the line joining the charges, its x-component \(F_{13x} = F_{13} \cdot \frac{{3cm}}{{5cm}}\) and y-component \(F_{13y} = F_{13} \cdot \frac{{4cm}}{{5cm}}\). The force \(F_{23}\) is along the y-direction, there is no x-component, and the y-component is equal to the force's magnitude \(F_{23y} = F_{23}\).
03

Calculate total force components

Superimpose the force vectors by adding their components. The total x-component of force \(F_{xTOTAL} = F_{13x}\) (since \(F_{23}\) has no x-component), and the total y-component of force \(F_{yTOTAL} = F_{13y} + F_{23y}\). Calculate these to find total forces in x and y directions.
04

Find magnitude and direction of the total force

To find the magnitude of the total force exerted on the 5nC charge calculate using Pythagoras' theorem: \(F_{TOTAL} = \sqrt{{F_{xTOTAL}^2 + F_{yTOTAL}^2}}\). For the direction with respect to the x-axis, apply trigonometry to get \(\Theta = tan^{-1} \left(\frac{{F_{yTOTAL}}}{{F_{xTOTAL}}\right)\)

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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 fundamentally describes how charges interact with each other. It is the foundational principle explaining the force between two point charges. The formula, \( F = k \times \frac{{|q_1 \times q_2|}}{{r^2}} \), is straightforward yet potent.
Coulomb's constant \( k \) is approximately \( 8.99 \times 10^9\, \text{N} \cdot \text{m}^2/\text{C}^2 \), representing the force exerted per unit charge squared per square meter. It allows us to scale the magnitude of electrical forces to practical units.
Here, \( q_1 \) and \( q_2 \) are the charges involved, and \( r \) is the distance between them. This calculation tells us the force magnitude, which is always along the line joining the two charges, implying it is also inherently directional.
  • Coulomb's Law helps us determine individual forces between charges to analyze their resultant effects when multiple charges act simultaneously.
Vector Components
In physics, understanding forces involves not just their magnitude but also their direction. This is where vector components come in. Every force acting in 2D or 3D space has parts that extend along each axis – most commonly, along the x and y-directions.
The concept of vector components allows decomposition of forces into more manageable parts. For instance, the force \( F_{13} \) exerted from the origin's charge to the point \( (3\, \text{cm}, 4\, \text{cm}) \) can be split into x- and y-components.
These components can be calculated as:
  • The x-component, \( F_{13x} = F_{13} \cdot \frac{{3\, \text{cm}}}{{5\, \text{cm}}} \)
  • The y-component, \( F_{13y} = F_{13} \cdot \frac{{4\, \text{cm}}}{{5\, \text{cm}}} \)
The fact that each vector component represents an aspect of the force in cases like this makes resolving the full force into its contributing directions essential to solve problems down to the axes of the coordinate system.
Force Magnitude and Direction
To fully characterize a two-dimensional force, one must identify both its magnitude and its direction. After determining the force components, combining them provides us with the complete picture.
The total force magnitude can be derived using the Pythagorean theorem. For our scenario, it is calculated as:
  • \( F_{\text{TOTAL}} = \sqrt{ F_{\text{xTOTAL}}^2 + F_{\text{yTOTAL}}^2} \)
This value represents the overall effect of all contributing forces exerted on the charge.
Direction is equally important. It is generally expressed in degrees from a significant axis, typically the x-axis. Calculating this angle uses the tangent inverse function:
  • \( \Theta = \tan^{-1} \left( \frac{F_{\text{yTOTAL}}}{F_{\text{xTOTAL}}} \right) \)
This direction aids in visualizing the force's inclination with respect to the positive x-direction, making interpretation in a coordinate system consistently intuitive.

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Most popular questions from this chapter

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.)

Two \(1.20 \mathrm{~m}\) nonconducting rods meet at a right angle. One rod carries \(+2.50 \mu \mathrm{C}\) of charge distributed uniformly along its length, and the other carries \(-2.50 \mu \mathrm{C}\) distributed uniformly along it (Fig. \(\mathbf{P} 2 \mathbf{1 . 8 5}\) ). (a) Find the magnitude and direction of the electric field these rods produce at point \(P,\) which is \(60.0 \mathrm{~cm}\) from each rod. (b) If an electron is released at \(P\), what are the magnitude and direction of the net force that these rods exert on it?

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.

Two small aluminum spheres, each having mass \(0.0250 \mathrm{~kg}\), are separated by \(80.0 \mathrm{~cm}\). (a) How many electrons does each sphere contain? (The atomic mass of aluminum is \(26.982 \mathrm{~g} / \mathrm{mol}\), and its atomic number is \(13 .\) ) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude \(1.00 \times 10^{4} \mathrm{~N}\) (roughly 1 ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

A point charge is placed at each corner of a square with side length a. All charges have magnitude \(q\). Two of the charges are positive and two are negative (Fig. E21.38). What is the direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of \(q\) and \(a\) ?

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