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1) Point charges q, = 75 nC, q, = -36 nC, and (a) Find the electric field at P. (b) Find the magnitude of the electric field at P. (c) Find the direction of the electric field at P. (d) Find the potential at P. (e) Find the electrical force on the charge q,. (k = 9×10° N×m²/C², and n= 93 16 nC are arranged in Figure. 10-9) 91 = 75 nC 92 = -36 nC + 3 m 3 m → X 4 m 93 = 16 nC

1) Point charges q, = 75 nC, q, = -36 nC, and (a) Find the electric field at P. (b) Find the magnitude of the electric field at P. (c) Find the direction of the electric field at P. (d) Find the potential at P. (e) Find the electrical force on the charge q,. (k = 9×10° N×m²/C², and n= 93 16 nC are arranged in Figure. 10-9) 91 = 75 nC 92 = -36 nC + 3 m 3 m → X 4 m 93 = 16 nC

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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16 nC are arranged in Figure. 1) Point charges q = 75 nC, q, = -36 nC, and q3 (a) Find the electric field at P. (b) Find the magnitude of the electric field at P. (c) Find the direction of the electric field at P. (d) Find the potential at P. (e) Find the electrical force on the charge q,. (k = 9×10° N×m²/C², and n = 10-9) y 91 = 75 nC 92 = -36 nC 3 m 3 m + →X 4 m 93 = 16 nC
Transcribed Image Text:16 nC are arranged in Figure. 1) Point charges q = 75 nC, q, = -36 nC, and q3 (a) Find the electric field at P. (b) Find the magnitude of the electric field at P. (c) Find the direction of the electric field at P. (d) Find the potential at P. (e) Find the electrical force on the charge q,. (k = 9×10° N×m²/C², and n = 10-9) y 91 = 75 nC 92 = -36 nC 3 m 3 m + →X 4 m 93 = 16 nC
2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K = 3 and K, = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m². The distance is d = 0.05 m. A potential difference of AV = 24 V is applied to the circuit. (a) Find the equivalent capacitance of the system. (b) Find the potential AV1. (E,=8.85×10-12 C²/N×m²) K1 A/2 AV1 K2 A/2 d K1 K1 A/2 2d/3 d/3 K2 A/2 d AV = 24 V F-----
Transcribed Image Text:2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K = 3 and K, = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m². The distance is d = 0.05 m. A potential difference of AV = 24 V is applied to the circuit. (a) Find the equivalent capacitance of the system. (b) Find the potential AV1. (E,=8.85×10-12 C²/N×m²) K1 A/2 AV1 K2 A/2 d K1 K1 A/2 2d/3 d/3 K2 A/2 d AV = 24 V F-----
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