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In the diagram, q1 = +6.39 • 109 C and q2 = +3.22. 10-⁹ C. What is the electric field at point P? Include a + or - sign to indicate the direction. P Eqz 91 (+) Eaz 0.424 m 0.636 m 92 (+)

In the diagram, q1 = +6.39 • 109 C and q2 = +3.22. 10-⁹ C. What is the electric field at point P? Include a + or - sign to indicate the direction. P Eqz 91 (+) Eaz 0.424 m 0.636 m 92 (+)

College Physics
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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In the diagram, \( q_1 = +6.39 \times 10^{-9} \, \text{C} \) and \( q_2 = +3.22 \times 10^{-9} \, \text{C} \). What is the electric field at point P? Include a + or - sign to indicate the direction. Diagram Explanation: - A horizontal line connects charge \( q_1 \) on the left, point P in the center, and charge \( q_2 \) on the right. - \( q_1 \) is positioned 0.424 meters to the left of point P. - \( q_2 \) is positioned 0.636 meters to the right of point P. - The electric field due to \( q_1 \) at point P (\( E_{q1} \)) is shown pointing to the right (positive direction). - The electric field due to \( q_2 \) at point P (\( E_{q2} \)) is shown pointing to the left (negative direction). (Remember, negative electric forces (-E) will point left, positive electric forces (+E) will point right.)
Transcribed Image Text:In the diagram, \( q_1 = +6.39 \times 10^{-9} \, \text{C} \) and \( q_2 = +3.22 \times 10^{-9} \, \text{C} \). What is the electric field at point P? Include a + or - sign to indicate the direction. Diagram Explanation: - A horizontal line connects charge \( q_1 \) on the left, point P in the center, and charge \( q_2 \) on the right. - \( q_1 \) is positioned 0.424 meters to the left of point P. - \( q_2 \) is positioned 0.636 meters to the right of point P. - The electric field due to \( q_1 \) at point P (\( E_{q1} \)) is shown pointing to the right (positive direction). - The electric field due to \( q_2 \) at point P (\( E_{q2} \)) is shown pointing to the left (negative direction). (Remember, negative electric forces (-E) will point left, positive electric forces (+E) will point right.)
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