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Chapter 21: Problem 44

Charge \(q_{1}=1.4 \cdot 10^{-8} \mathrm{C}\) is placed at the origin. Charges \(q_{2}=-1.8 \cdot 10^{-8} \mathrm{C}\) and \(q_{3}=2.1 \cdot 10^{-8} \mathrm{C}\) are placed at points \((0.18 \mathrm{~m}, 0 \mathrm{~m})\) and \((0 \mathrm{~m}, 0.24 \mathrm{~m}),\) respec- tively, as shown in the figure. Determine the net electrostatic force (magnitude and direction) on charge \(q_{3}\)

Short Answer

Expert verified
The net electrostatic force on charge \(q_{3}\) has a magnitude of approximately \(3.46 \times 10^{-7} N\) and a direction of approximately \(-111.77°\) from the x-axis.

Step by step solution

01

Determine magnitude of force between \(q_{1}\) and \(q_{3}\)

Coulomb's Law states that the magnitude of the electrostatic force F between two charges \(q_{1}\) and \(q_{2}\) is given by: \(F=\frac{k \cdot |q_{1} \cdot q_{2}|}{r^2}\) Where k is the electrostatic constant, \(\approx 8.99 \times 10^{9} Nm^2/C^2\). The distance between charges \(q_{1}\) and \(q_{3}\) is equal to the distance from the origin to \((0, 0.24m)\): \(r=\sqrt{0^2 + (0.24)^2}=0.24m\) Now, we can plug in the given values to find the force magnitude: \(F_{13}=\frac{8.99 \times 10^{9} Nm^2/C^2 \cdot |1.4 \times 10^{-8} C \cdot 2.1 \times 10^{-8} C|}{(0.24m)^2} \approx 1.82 \times 10^{-7}N\)
02

Determine magnitude of force between \(q_{2}\) and \(q_{3}\)

Follow the same procedure as step 1, but for charges \(q_{2}\) and \(q_{3}\). The distance between charges \(q_{2}\) and \(q_{3}\) is: \(r=\sqrt{(0.18)^2 + (0.24)^2} \approx 0.30 m\) Now, calculate the force magnitude: \(F_{23}=\frac{8.99 \times 10^{9} Nm^2/C^2 \cdot |-1.8 \times 10^{-8} C \cdot 2.1 \times 10^{-8} C|}{(0.30m)^2} \approx 1.68 \times 10^{-7} N\)
03

Find the components of the forces \(F_{13}\) and \(F_{23}\)

In order to add the forces as vectors, we need to find their respective x and y components. For \(F_{13}\), the force acts directly upward, so: \(F_{13x} = 0\) \(F_{13y} = 1.82 \times 10^{-7} N\) Now, we need to find the components of \(F_{23}\). Find the angle between the horizontal line and the line connecting charges \(q_{2}\) and \(q_{3}\): \(\theta = \arctan(\frac{0.24}{0.18}) \approx 53.13°\) Now, find the x and y components of \(F_{23}\) using trigonometry: \(F_{23x} = F_{23} \cos{\theta} \approx -1.28 \times 10^{-7} N\) \(F_{23y} = F_{23} \sin{\theta} \approx 1.39 \times 10^{-7} N\)
04

Calculate the net force on \(q_{3}\)

Now, add the x and y components of the forces to find the net force: \(F_{netx} = F_{13x} + F_{23x} = 0 + (-1.28 \times 10^{-7} N) = -1.28 \times 10^{-7} N\) \(F_{nety} = F_{13y} + F_{23y} = 1.82 \times 10^{-7} N + 1.39 \times 10^{-7} N = 3.21 \times 10^{-7} N\) The net force's magnitude is found using the Pythagorean theorem: \(F_{net} = \sqrt{(-1.28 \times 10^{-7} N)^2 + (3.21 \times 10^{-7} N)^2} \approx 3.46 \times 10^{-7} N\)
05

Calculate the direction of the net force

Now, we can find the direction of the net force: \(\theta_{net} = \arctan(\frac{F_{nety}}{F_{netx}}) = \arctan(\frac{3.21 \times 10^{-7} N}{-1.28 \times 10^{-7} N}) \approx -111.77°\) Since the angle is negative, it means the net force's direction is in the 4th quadrant. The net electrostatic force on charge \(q_{3}\) has a magnitude of approximately \(3.46 \times 10^{-7} N\) and a direction of approximately \(-111.77°\) from the x-axis.

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Most popular questions from this chapter

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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