Calculate the resultant electric field strength and direction at point P. P(1.5m , 2m)q2= +20 µC92= +10 nCx= 1 mx= 3 mE(N/C)Angle ( °)x-comp(N/C)y-comp(N/C)E1 =E2=Ex=Ey=NET electric chargedirection/angle

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Answered: (1.5m , 2m) qq= +20 µC q2= +10 nC x= 1…

(1.5m , 2m) qq= +20 µC q2= +10 nC x= 1 m X= 3 m E(N/C) Angle ( º) X-comp(N/C) y-comp(N/C) E1 = E2= Ex= Ey= NET electric charge direction/angle

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter23: Continuous Charge Distributions And Gauss's Law

Section: Chapter Questions

Problem 38AP

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Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

Calculate the resultant electric field strength and direction at point P.

P
(1.5m , 2m)
q2= +20 µC
92= +10 nC
x= 1 m
x= 3 m
E(N/C)
Angle ( °)
x-comp(N/C)
y-comp(N/C)
E1 =
E2=
Ex=
Ey=
NET electric charge
direction/angle

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