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Chapter 16: Problem 78

A point charge \(q_{1}=+5.0 \mu \mathrm{C}\) is fixed in place at \(x=0\) and a point charge \(q_{2}=-3.0 \mu \mathrm{C}\) is fixed at \(x=\) $-20.0 \mathrm{cm} .\( Where can we place a point charge \)q_{3}=-8.0 \mu \mathrm{C}$ so that the net electric force on \(q_{1}\) due to \(q_{2}\) and \(q_{3}\) is zero?

Short Answer

Expert verified
Answer: The point charge \(q_3\) should be placed at \(x = 20.8\,cm\).

Step by step solution

01

Write down the known quantities

We are given: - Point charge \(q_1 = +5.0\,\mu C\) - Position of \(q_1\): \(x = 0\) - Point charge \(q_2 = -3.0\,\mu C\) - Position of \(q_2\): \(x = -20.0\,cm\) - Point charge \(q_3 = -8.0\,\mu C\) - Goal: Find the position of \(q_3\) such that the net electric force on \(q_1\) is zero.
02

Define the force relation between the charges

Let's consider the following distances: - \(d_{12}\): distance between \(q_1\) and \(q_2\) - \(d_{13}\): distance between \(q_1\) and \(q_3\) We are looking for the position of \(q_3\), so the force exerted on \(q_1\) by \(q_2\) is equal and opposite to the force exerted on \(q_1\) by \(q_3\): $$F_{12} = F_{13}$$
03

Apply Coulomb's Law for both forces

Coulomb's Law states that the force between two point charges is given by: $$F = k\frac{|q_1||q_2|}{d^2}$$ where \(k = 8.99 \times 10^9 \,\text{N}\,\text{m}^2\,\text{C}^{-2}\) is the electrostatic constant. We can now use this formula to write down the equations for the forces between the charges: $$F_{12} = k\frac{|q_1||q_2|}{(d_{12})^2}$$ $$F_{13} = k\frac{|q_1||q_3|}{(d_{13})^2}$$ Since \(F_{12} = F_{13}\), we can write: $$k\frac{|q_1||q_2|}{(d_{12})^2} = k\frac{|q_1||q_3|}{(d_{13})^2}$$
04

Simplify and solve for the position of \(q_3\)

We can simplify the last equation, and solve for \(d_{13}\): $$\frac{|q_2|}{(d_{12})^2} = \frac{|q_3|}{(d_{13})^2}$$ $$d_{13} = \sqrt{\frac{|q_3|}{|q_2|}(d_{12})^2}$$ The distance between the \(q_1\) and \(q_2\) is given by \(d_{12} = |x_{q_1} - x_{q_2}| = |-(-20.0\,cm)| = 20.0\,cm\). We can now use the values of the charges and the distance \(d_{12}\) to find the distance \(d_{13}\): $$d_{13} = \sqrt{\frac{|-8.0\,\mu C|}{|-3.0\,\mu C|}(20.0\,cm)^2} = \sqrt{\frac{8.0}{3.0}(20.0\,cm)^2} \approx 20.8\,cm$$
05

Determine the location of \(q_3\)

Since \(q_3\) should be placed on the \(x\)-axis, we have to add the distance \(d_{13}\) to the \(x\)-coordinate of \(q_1\), which is \(x_{q_1} = 0\). This gives: $$x_{q_3} = x_{q_1} + d_{13} = 20.8\,cm$$ Therefore, the point charge \(q_3\) should be placed at \(x = 20.8\,cm\) to make the net electric force on \(q_1\) due to \(q_2\) and \(q_3\) equal to zero.

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Most popular questions from this chapter

Suppose a charge \(q\) is placed at point \(x=0, y=0 .\) A second charge \(q\) is placed at point \(x=8.0 \mathrm{m}, y=0 .\) What charge must be placed at the point \(x=4.0 \mathrm{m}, y=0\) in order that the field at the point $x=4.0 \mathrm{m}, y=3.0 \mathrm{m} \mathrm{be}$ zero?
In fair weather, over flat ground, there is a downward electric field of about \(150 \mathrm{N} / \mathrm{C} .\) (a) Assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is $150 \mathrm{N} / \mathrm{C}$ pointing radially inward, calculate the total charge on the Earth and the charge per unit area. (b) At an altitude of $250 \mathrm{m}\( above Earth's surface, the field is only \)120 \mathrm{N} / \mathrm{C}$ Calculate the charge density (charge per unit volume) of the air (assumed constant). [Hint: See Conceptual Example \(16.8 .]\)
A positively charged rod is brought near two uncharged conducting spheres of the same size that are initially touching each other (diagram a). The spheres are moved apart and then the charged rod is removed (diagram b). (a) What is the sign of the net charge on sphere 1 in diagram b? (b) In comparison with the charge on sphere \(1,\) how much and what sign of charge is on sphere \(2 ?\)
Two tiny objects with equal charges of \(7.00 \mu \mathrm{C}\) are placed at the two lower corners of a square with sides of \(0.300 \mathrm{m},\) as shown. Find the electric field at point \(A\) the upper left corner.
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