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Chapter 21: Problem 66

Three point charges are positioned on the \(x\) -axis: \(+64.0 \mu \mathrm{C}\) at \(x=0.00 \mathrm{~cm},+80.0 \mu \mathrm{C}\) at \(x=25.0 \mathrm{~cm},\) and \(-160.0 \mu C\) at \(x=50.0 \mathrm{~cm} .\) What is the magnitude of the electrostatic force acting on the \(+64.0-\mu C\) charge?

Short Answer

Expert verified
Answer: The magnitude of the electrostatic force acting on the +64.0 µC charge is 6.8 N.

Step by step solution

01

Identify the charges and their positions

We have three point charges: 1. \(q_1 = +64.0 \ \mu C\) at \(x_1=0.00 \ cm\) 2. \(q_2 = +80.0 \ \mu C\) at \(x_2=25.0 \ cm\) 3. \(q_3 = -160.0 \ \mu C\) at \(x_3=50.0 \ cm\) Our goal is to find the magnitude of the electrostatic force acting on the \(q_1\) charge.
02

Use Coulomb's Law to calculate the forces

Coulomb's Law states that the magnitude of the force between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by: \(F = k \frac{|q_1 \cdot q_2|}{r^2}\) Where \(k\) is the electrostatic constant, \(k = 8.99 \times 10^9 \ N \cdot m^2 \ / \ C^2\). Now, we will calculate the force between each pair of charges.
03

Step 2a: Calculate the force between \(q_1\) and \(q_2\)

The distance between \(q_1\) and \(q_2\) is \(r_{12} = x_2 - x_1 = 25.0 \ cm = 0.25 \ m\). Thus, the force between \(q_1\) and \(q_2\) is: \(F_{12} = k \frac{|q_1 \cdot q_2|}{r^2_{12}} = 8.99 \times 10^9 \frac{|(+64 \times 10^{-6}) \cdot (+80 \times 10^{-6})|}{(0.25)^2\ m^2} = 11.5 \ N\) Since both charges are positive, the force is repulsive, so \(F_{12}\) acts to the right.
04

Step 2b: Calculate the force between \(q_1\) and \(q_3\)

The distance between \(q_1\) and \(q_3\) is \(r_{13} = x_3 - x_1 = 50.0 \ cm = 0.50 \ m\). Thus, the force between \(q_1\) and \(q_3\) is: \(F_{13} = k \frac{|q_1 \cdot q_3|}{r^2_{13}} = 8.99 \times 10^9 \frac{|(+64 \times 10^{-6}) \cdot (-160 \times 10^{-6})|}{(0.50)^2\ m^2} = 18.3 \ N\) Since the charges have opposite signs, the force is attractive, so \(F_{13}\) acts to the left.
05

Calculate the net force on \(q_1\)

To find the net force acting on the \(+64.0\ \mu C\) charge, we simply add the forces resulting from the \(+80.0\ \mu C\) and the \(-160.0\ \mu C\) charges in terms of their directions. \(F_{net} = F_{13} + (-F_{12}) = 18.3 - 11.5 = 6.8 \ N\) Thus, the magnitude of the electrostatic force acting on the \(+64.0\ \mu C\) charge is \(6.8\ N\).

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Most popular questions from this chapter

A particle (charge \(=+19.0 \mu C)\) is located on the \(x\) -axis at \(x=-10.0 \mathrm{~cm},\) and a second particle (charge \(=-57.0 \mu \mathrm{C})\) is placed on the \(x\) -axis at \(x=+20.0 \mathrm{~cm} .\) What is the magnitude of the total electrostatic force on a third particle (charge = \(-3.80 \mu \mathrm{C})\) placed at the origin \((x=0) ?\)

A negative charge, \(-q\), is fixed at the coordinate (0,0) It is exerting an attractive force on a positive charge, \(+q,\) that is initially at coordinate \((x, 0)\). As a result, the positive charge accelerates toward the negative charge. Use the binomial expansion \((1+x)^{n} \approx 1+n x,\) for \(x \ll 1,\) to show that when the positive charge moves a distance \(\delta \ll x\) closer to the negative charge, the force that the negative charge exerts on it increases by \(\Delta F=2 k q^{2} \delta / x^{3}\) .

Two point charges are fixed on the \(x\) -axis: \(q_{1}=6.0 \mu \mathrm{C}\) is located at the origin, \(O,\) with \(x_{1}=0.0 \mathrm{~cm},\) and \(q_{2}=-3.0 \mu \mathrm{C}\) is located at point \(A,\) with \(x_{2}=8.0 \mathrm{~cm} .\) Where should a third charge, \(q_{3},\) be placed on the \(x\) -axis so that the total electrostatic force acting on it is zero? a) \(19 \mathrm{~cm}\) c) \(0.0 \mathrm{~cm}\) e) \(-19 \mathrm{~cm}\) b) \(27 \mathrm{~cm}\) d) \(8.0 \mathrm{~cm}\)

In gaseous sodium chloride, chloride ions have one more electron than they have protons, and sodium ions have one more proton than they have electrons. These ions are separated by about \(0.24 \mathrm{nm}\). Suppose a free electron is located \(0.48 \mathrm{nm}\) above the midpoint of the sodium chloride molecule. What are the magnitude and the direction of the electrostatic force the molecule exerts on it?

Your sister wants to participate in the yearly science fair at her high school and asks you to suggest some exciting project. You suggest that she experiment with your recently created electron extractor to suspend her cat in the air. You tell her to buy a copper plate and bolt it to the ceiling in her room and then use your electron extractor to transfer electrons from the plate to the cat. If the cat weighs \(7.00 \mathrm{~kg}\) and is suspended \(2.00 \mathrm{~m}\) below the ceiling, how many electrons have to be extracted from the cat? Assume that the cat and the metal plate are point charges.
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