Step 1 The strategy to use when calculating the force exerted by a set of several point charges is to find first the forces exerted by each of the point charges individually. The force exerted by the entire set of charges is the net force resulting from the forces exerted by the individual charges. The forces from the individual charges in this problem are shown in the diagram below. F12 191 0.100_ m Fg F13 92 The charge q, exerts a repulsive force F,, on q, as shown in the diagram. The distance separating these 12 charges is r,, = 0.306 m so the magnitude of this repulsive force is given by the following equation where k. is the Coulomb constant in SI units. 19,|1921 ke 12 12 28.98 V 29 x 10-18 = (8.99 x 10° N •m²/c?y 0.306 V 0.306 m) 2.78 V 2.78 x 10-6 N Step 2 The negative charge g. exerts an attractive force F 13 on the positive charge q.. The distance between these two charges is r,, = 0.100 m, so we have F13 = ke 13 14.49 14.5 x 10-18 = (8.99 x 10° N :m/c?) 0.100 | )² 0.1 m 1.30 V 1.3 x 10-5 N. Step 3 The magnitude of the force FR resulting from F12 and F13 can be found by using the Pythagorean theorem. From the diagram we see that the x-component of the resultant force is F. = F., and its y-component is R, x = F,3, so that FR, Y FR = V (FR, + (FR, = v V 1.88 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in y 1.371 Your response is within 10% the correct value. This may be due to roundoff error, or you could have a mistake in your Submit Skip (you cannot come back)
Step 1 The strategy to use when calculating the force exerted by a set of several point charges is to find first the forces exerted by each of the point charges individually. The force exerted by the entire set of charges is the net force resulting from the forces exerted by the individual charges. The forces from the individual charges in this problem are shown in the diagram below. F12 191 0.100_ m Fg F13 92 The charge q, exerts a repulsive force F,, on q, as shown in the diagram. The distance separating these 12 charges is r,, = 0.306 m so the magnitude of this repulsive force is given by the following equation where k. is the Coulomb constant in SI units. 19,|1921 ke 12 12 28.98 V 29 x 10-18 = (8.99 x 10° N •m²/c?y 0.306 V 0.306 m) 2.78 V 2.78 x 10-6 N Step 2 The negative charge g. exerts an attractive force F 13 on the positive charge q.. The distance between these two charges is r,, = 0.100 m, so we have F13 = ke 13 14.49 14.5 x 10-18 = (8.99 x 10° N :m/c?) 0.100 | )² 0.1 m 1.30 V 1.3 x 10-5 N. Step 3 The magnitude of the force FR resulting from F12 and F13 can be found by using the Pythagorean theorem. From the diagram we see that the x-component of the resultant force is F. = F., and its y-component is R, x = F,3, so that FR, Y FR = V (FR, + (FR, = v V 1.88 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in y 1.371 Your response is within 10% the correct value. This may be due to roundoff error, or you could have a mistake in your Submit Skip (you cannot come back)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Step 1**
The strategy to use when calculating the force exerted by a set of several point charges is to find first the forces exerted by each of the point charges individually. The force exerted by the entire set of charges is the net force resulting from the forces exerted by the individual charges. The forces from the individual charges in this problem are shown in the diagram below.
**Diagram Description:**
- Three charges are shown: \( q_1 \), \( q_2 \), and \( q_3 \).
- \( q_1 \) is at the origin of the coordinate system.
- The distance between \( q_1 \) and \( q_2 \) is 0.306 m.
- The force \( \vec{F}_{12} \) is directed from \( q_2 \) to \( q_1 \).
- The distance between \( q_1 \) and \( q_3 \) is 0.100 m.
- The force \( \vec{F}_{13} \) is directed from \( q_3 \) to \( q_1 \).
The charge \( q_2 \) exerts a repulsive force \( \vec{F}_{12} \) on \( q_1 \). The magnitude of this repulsive force is given by the equation:
\[ F_{12} = k_e \frac{|q_1||q_2|}{r_{12}^2} \]
Calculating:
\[
F_{12} = (8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \times \frac{(29 \times 10^{-18} \, \text{C}^2)}{(0.306 \, \text{m})^2} = 2.78 \times 10^{-6} \, \text{N}
\]
**Step 2**
The negative charge \( q_3 \) exerts an attractive force \( \vec{F}_{13} \) on the positive charge \( q_1 \). We have:
\[
F_{13} = k_e \frac{|q_1||q_3|}{r_{13}^2}
\]
Calculating:
\[
F_{13} = (8.99 \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F600b05a1-578f-4b99-b7bd-f475fff2bde2%2F8f8e0412-f390-4ce4-8798-76e96fcb1542%2F3l59hp_processed.png&w=3840&q=75)
Transcribed Image Text:**Step 1**
The strategy to use when calculating the force exerted by a set of several point charges is to find first the forces exerted by each of the point charges individually. The force exerted by the entire set of charges is the net force resulting from the forces exerted by the individual charges. The forces from the individual charges in this problem are shown in the diagram below.
**Diagram Description:**
- Three charges are shown: \( q_1 \), \( q_2 \), and \( q_3 \).
- \( q_1 \) is at the origin of the coordinate system.
- The distance between \( q_1 \) and \( q_2 \) is 0.306 m.
- The force \( \vec{F}_{12} \) is directed from \( q_2 \) to \( q_1 \).
- The distance between \( q_1 \) and \( q_3 \) is 0.100 m.
- The force \( \vec{F}_{13} \) is directed from \( q_3 \) to \( q_1 \).
The charge \( q_2 \) exerts a repulsive force \( \vec{F}_{12} \) on \( q_1 \). The magnitude of this repulsive force is given by the equation:
\[ F_{12} = k_e \frac{|q_1||q_2|}{r_{12}^2} \]
Calculating:
\[
F_{12} = (8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \times \frac{(29 \times 10^{-18} \, \text{C}^2)}{(0.306 \, \text{m})^2} = 2.78 \times 10^{-6} \, \text{N}
\]
**Step 2**
The negative charge \( q_3 \) exerts an attractive force \( \vec{F}_{13} \) on the positive charge \( q_1 \). We have:
\[
F_{13} = k_e \frac{|q_1||q_3|}{r_{13}^2}
\]
Calculating:
\[
F_{13} = (8.99 \times
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