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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

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) \(8.0 \mathrm{~cm}\)

A positive charge \(q_{1}=1.00 \mu \mathrm{C}\) is fixed at the origin, and a second charge \(q_{2}=-2.00 \mu \mathrm{C}\) is fixed at \(x=10.0 \mathrm{~cm} .\) Where along the \(x\) -axis should a third charge be positioned so that it experiences no force?

Another unit of charge is the electrostatic unit (esu). It is defined as follows: Two point charges, each of 1 esu and separated by \(1 \mathrm{~cm}\), exert a force of exactly 1 dyne on each other: 1 dyne \(=1 \mathrm{~g} \mathrm{~cm} / \mathrm{s}^{2}=1 \cdot 10^{-5} \mathrm{~N}\). a) Determine the relationship between the esu and the coulomb. b) Determine the relationship between the esu and the elementarv charoe

The force between a charge of \(25 \mu C\) and a charge of \(-10 \mu C\) is \(8.0 \mathrm{~N}\). What is the separation between the two charges? a) \(0.28 \mathrm{~m}\) c) \(0.45 \mathrm{~m}\) b) \(0.53 \mathrm{~m}\) d) \(0.15 \mathrm{~m}\)

Suppose the Earth and the Moon carried positive charges of equal magnitude. How large would the charge need to be to produce an electrostatic repulsion equal to \(1.00 \%\) of the gravitational attraction between the two bodies?
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