/*! 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 14 A charge of \(-3.00 \mathrm{nC}\... [FREE SOLUTION] | 91Ó°ÊÓ

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A charge of \(-3.00 \mathrm{nC}\) and a charge of \(-5.80 \mathrm{nC}\) are separated by a distance of \(50.0 \mathrm{~cm}\). Find the position at which a third charge of \(+7.50 \mathrm{nC}\) can be placed so that the net electrostatic force on it is zero.

Short Answer

Expert verified
After going through steps 1, 2, and 3, you will get two positions for the \(+7.50 \mathrm{nC}\) charge, so that the net electrostatic force on it will be zero. One position is in between the two charges, and the other position is to the right of both charges.

Step by step solution

01

Understanding the scenario

There are two possibilities for where the third charge (\(Q_3=+7.50 \mathrm{nC}\)) could be placed in relation to the other two charges (\(Q_1=-3.00 \mathrm{nC}\), \(Q_2=-5.80 \mathrm{nC}\)): either between the two charges, or on either side of the two charges. Each possibility will give a different position, and both positions must be calculated.
02

Applying Coulomb's law

For the scenario where \(Q_3\) is between \(Q_1\) and \(Q_2\), assuming \(x\) as the distance from \(Q_3\) to \(Q_1\), the net force equation would be \(\frac{kQ_1Q_3}{x^2}=\frac{kQ_2Q_3}{(0.5-x)^2}\). For the scenario where \(Q_3\) is outside \(Q_1\) and \(Q_2\), the net force equation would be \(\frac{kQ_1Q_3}{x^2}+\frac{kQ_2Q_3}{(0.5+x)^2}=0\). Note that k is the electrostatic constant.
03

Solving the equations

By solving these equations, you can find the two possible values for \(x\), those are the two positions at which the third charge can be placed so that the net electrostatic force on it is zero. This involves simplifying the equation for each scenario by crossing out the same terms, rearranging it and then solving it using algebra and root-finding methods.

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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 of electrostatics that describes the interaction between electric charges. It states that the force between two point charges is directly proportional to the product of the magnitudes of the charges, and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
\[F = k \frac{|Q_1 Q_2|}{r^2}\]where:
  • \(F\) is the force between the charges,
  • \(k\) is Coulomb's constant (\(8.9875 \times 10^9 \text{ Nm}^2/\text{C}^2\)),
  • \(Q_1\) and \(Q_2\) are the magnitudes of the charges, and
  • \(r\) is the distance between the centers of the two charges.
Coulomb's Law is essential for understanding how charged objects interact. It is crucial in calculating the electrostatic forces and plays a significant role in determining equilibrium positions, such as in this exercise.

When applied to multiple charge systems, as seen in the exercise, the superposition principle is used. This means you sum the forces exerted on a charge by other charges to find the net force acting on it.
Electric Charge
Electric charge is a fundamental property of matter, appearing in two types: positive and negative. Objects with the same type of charge repel each other, while those with opposite charges attract each other.
The unit of charge is the Coulomb (C), a measure of the amount of electricity transported by a steady current of one ampere in one second. In this problem, charges are described in nanocoulombs (nC), which is \(10^{-9}\) Coulombs.

The problem presents three charges: two negative and one positive. It's important to note that the negative charges will repel the positive one. Thus, our goal when placing the third charge is to find a position where these repulsive forces cancel out, creating a net electrostatic force of zero.
  • A negative charge such as \(-3.00 \text{ nC}\) contributes to pushing the positive charge away.
  • Similarly, the \(-5.80 \text{ nC}\) charge will also exert a repelling force.
  • Understanding interactions between these charges helps predict the behavior and outcome of the forces at play.
Algebraic Solutions
In the context of this exercise, algebra is used to solve equations derived from Coulomb's Law to find equilibrium positions. We aim to determine the position \(x\), where the third charge can be placed, such that the net electrostatic force is zero.

This involves setting up expressions for the forces and solving for \(x\). When the third charge is placed between the negative charges, we equate the electrostatic forces due to each charge to each other because their net force must be zero for equilibrium.
  • When calculating where \(Q_3\) should be set, let \(x\) be the distance between \(Q_3\) and \(Q_1\).
  • Here, the equation \(\frac{kQ_1Q_3}{x^2} = \frac{kQ_2Q_3}{(0.5-x)^2}\) is arranged and simplified algebraically.
After formulating this equation, terms like the electrostatic constant \(k\) and the charge of \(Q_3\) are cancelled out since both sides are equal.

Next, solving involves rearranging the equation to isolate \(x\) and using techniques such as square-root extraction or quadratic formulas to find the two possible solutions for \(x\). This approach ensures that all mathematical steps are logically sound towards reaching the solution of no net force condition.

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Most popular questions from this chapter

In the Millikan oil-drop experiment, an atomizer (a sprayer with a fine nozzle) is used to introduce many tiny droplets of oil between two oppositely charged parallel metal plates. Some of the droplets pick up one or more excess electrons. The charge on the plates is adjusted so that the electric force on the excess electrons exactly balances the weight of the droplet. The idea is to look for a droplet that has the smallest electric force and assume it has only one excess electron. This strategy lets the observer measure the charge on the electron. Suppose we are using an electric field of \(3 \times 10^{4} \mathrm{~N} / \mathrm{C}\). The charge on one electron is about \(1.6 \times 10^{-19} \mathrm{C}\). Estimate the radius of an oil drop of density \(858 \mathrm{~kg} / \mathrm{m}^{3}\) for which its weight could be balanced by the electric force of this field on one electron. (Problem 38 is courtesy of E. F. Redish. For more problems of this type, visit www.physics.umd.edu/perg/.)

An electron is accelerated by a constant electric field of magnitude \(300 \mathrm{~N} / \mathrm{C} .\) (a) Find the acceleration of the electron. (b) Use the equations of motion with constant acceleration to find the electron's speed after \(1.00 \times 10^{-8} \mathrm{~s}\), assuming it starts from rest.

Two \(2.0-\mathrm{g}\) spheres are suspended by \(10.0-\mathrm{cm}\) -long light strings (Fig. P15.57). A uniform electric field is applied in the \(x\) -direction. If the spheres have charges of \(-5.0 \times\) \(10^{-8}\) Cand \(+5.0 \times 10^{-8} \mathrm{C}\), determine the electric field intensity that enables the spheres to be in equilibrium at \(\theta=10^{\circ}\).

Two point charges lie along the y-axis. A charge of \(q_{1}=\) \(-9.0 \mu \mathrm{C}\) is at \(y=6.0 \mathrm{~m}\), and a charge of \(q_{2}=-8.0 \mu \mathrm{C}\) is at \(y=-4.0 \mathrm{~m}\). Locate the point (other than infinity) at which the total electric field is zero.

(a) Two identical point charges \(+q\) are located on the \(y\) -axis at \(y=+a\) and \(y=-a\). What is the electric field along the \(x\) -axis at \(x=b_{p}^{2}(b) A\) circular ring of charge of radius \(a\) has a total positive charge \(Q\) distributed uniformly around it. The ring is in the \(x=0\) plane with its center at the origin. What is the electric field along the \(x\) -axis at \(x=b\) due to the ring of charge? (Hint: Consider the charge \(Q\) to consist of many pairs of identical point charges positioned at the ends of diameters of the ring.)

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