
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Transcribed Image Text:Find the electric field, in rectangular component form, at the point, P in the diagram below due to the stationary point charges, 91 and q2,
where 91 = -2.00 HC and q2 = +1.00 pC. (Assume that k = 9.0 x 109 Nm2/c2)
y
X
91
5.0 cm
P
i12 cm
0 92
a. Ex = 8.2 x 107 N/C; Ey = -3.6x 107 N/C
b. Ex = -4.1 x 107 N/C; Ey = -3.6 x 107 N/C
c. Ex = -5.0 x 105 N/C ; Ey = 4.0 x 105 N/C
d. Ex = -4.1 x 105 N/C; Ey = -3.6 x 105 N/C
e. Ex = 6.7 x 105 N/C; Ey = 8.9 x 105 N/C
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- The figure below shows electric field generated by two charges q₁ and q2. What is most likely combination of 92 91 a) q1 = -3μC, q2= -12 μC b) q₁ = +6 μC, q2= +12 µC c) 9₁ = +2 μC, 92= -9 μC d) q1 = -8 μC, q2= +24 MC None of the answer choices is correct. Write your own 91 and 92?arrow_forwardA positive charge, 91 = = 5.75nC, is located at the origin (x = 0 m, y = 0 m). A negative charge, 92 - -2.85nC, is located on the x-axis at (x = 4 m, y = 0 m). Calculate the magnitude of the electric field at the location marked by the letter "X" on the figure below at (x = 4m, y = 3 m). = 4 Enter answer here 3 2 1 0 (9₁) 0 1 2 3 4 5 6 Image size: S M L Max Please enter a numerical answer below. Accepted formats are numbers or "e" based scientific notation e.g. 0.23, -2, 1e6, 5.23e-8 N/Carrow_forwardTwo point charges are situated in a cartesian coordinate system as follow; 91-9.6 x 10 C @ the coordinates (-0.49, 0.92) m 92= -3.4 x 106 C @ the coordinates (0.74, 0.45) m What is the direction of the net electric field at the origin (0, 0)? Answer this as an angle measured in degrees relative to the +x direction. Let answers in the first 3 quadrants be positive and negative in the 4th quadrant.arrow_forward
- Calculate the electric force that q3 exerts on q2. Then draw the force vector on the figure and label it F3on2.arrow_forwardFind the x and y components of the electric field produced by 91 and 92 in the figure shown below at point A and point B. (Take q₁ = 2.16 μC and q2 = -1.09 μC.) y, m 3 2 1 B 914 Point A Ex= Ey= Point B Ex = Ey = I 1 2 3 4 -x, marrow_forwardFind the net electric flux through the spherical closed surface shownin the figure below.arrow_forward
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