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Chapter 21: Problem 76

A point charge \(q_{1}=100 . \mathrm{nC}\) is at the origin of an \(x y\) -coordinate system, a point charge \(q_{2}=-80.0 \mathrm{nC}\) is on the \(x\) -axis at \(x=2.00 \mathrm{~m},\) and a point charge \(q_{3}=-60.0 \mathrm{nC}\) is on the \(y\) -axis at \(y=-2.00 \mathrm{~m} .\) Determine the net force (magnitude and direction) on \(q_{1}\).

Short Answer

Expert verified
Answer: The magnitude of the net force on \(q_1\) is \(22.5\:nN\), and the direction is \(36.9^\circ\) counterclockwise from the negative x-axis.

Step by step solution

01

Understanding the problem and setting up the variables

Let's first identify the given values and variables to set up the problem. The charges are: \(q_{1}=100\:nC\) (at the origin), \(q_{2}=-80.0\:nC\) (at \((2.00\:m,0)\)), and \(q_{3}=-60.0\:nC\) (at \((0,-2.00\:m)\)). We need to find the net force on \(q_1\), which is the sum of the individual forces acting on \(q_1\) due to \(q_2\) and \(q_3\).
02

Calculate the force between \(q_1\) and \(q_2\)

To calculate the force between \(q_1\) and \(q_2\), we can use Coulomb's Law which states: \(F=\frac{kq_1q_2}{r^2}\). In this case, the distance \(r_{12}\) between \(q_1\) (origin) and \(q_2\) (2.00 meters along the x-axis) is 2.00 meters. So, the force between \(q_1\) and \(q_2\) is given by: \(F_{12}=\frac{(9*10^9)(100*10^{-9})((-80)*10^{-9})}{(2)^2}=-18*10^{-9}N\) This force is acting in the negative x-direction.
03

Calculate the force between \(q_1\) and \(q_3\)

Now, we will calculate the force between \(q_1\) and \(q_3\) using Coulomb's Law, with the distance \(r_{13}\) being \(2.00\) meters. The force between \(q_1\) and \(q_3\) is given by: \(F_{13}=\frac{(9*10^9)(100*10^{-9})((-60)*10^{-9})}{(2)^2}=-13.5*10^{-9}N\) This force is acting in the positive y-direction.
04

Finding the net force on \(q_1\)

To find the net force on \(q_1\), we will sum the individual forces acting on \(q_1\) due to \(q_2\) and \(q_3\). The net force on \(q_1\) can be represented as a 2D vector: \(\vec{F}_{net} = \vec{F}_{12}+\vec{F}_{13}\) Given the direction of \(F_{12}\) (negative x-direction) and \(F_{13}\) (positive y-direction), we can rewrite \(\vec{F}_{net}\) as: \(\vec{F}_{net} = (-18*10^{-9}\hat{i}) + (-13.5*10^{-9}\hat{j})\)
05

Compute the magnitude and direction of the net force

We can compute the magnitude of the net force using the Pythagorean theorem: \(F_{net} = \sqrt{(-18*10^{-9})^2+(-13.5*10^{-9})^2}= 22.5*10^{-9}N\) Next, we need to find the angle between the net force and the negative x-axis. We can use the arctangent function for this purpose: \(\theta = \arctan\left(\frac{-13.5*10^{-9}}{-18*10^{-9}}\right)=36.9^\circ\) So, the net force magnitude on \(q_1\) is \(22.5\:nN\), and the direction is \(36.9^\circ\) counterclockwise from the negative x-axis.

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Most popular questions from this chapter

A current of \(5.00 \mathrm{~mA}\) is enough to make your muscles twitch. Calculate how many electrons flow through your skin if you are exposed to such a current for \(10.0 \mathrm{~s}\).

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The faraday is a unit of charge frequently encountered in electrochemical applications and named for the British physicist and chemist Michael Faraday. It consists of 1 mole of elementary charges. Calculate the number of coulombs in 1.000 faraday.
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