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g2 gl In the figure, particle 1 of charge q1-8.20x10-5 C and particle 2 of charge q2 . 3.28x10-C are fixed to an x axis, separated by a distance d 0.100 m. Calculate their net electric field E(x) as a function of x for the following positive and negative values of x, taking E to be positive when the vector E points to the right and negative when E points to the left What is E(-0.100)? 1.47x10 N/C You are correct. receipt no, is 161-1699 What is E(-0.0s0)? 5.904E8 N/C Tries 5/5 Previous Triss What is E(o.070)? 4349 N/C Tries 4/5 Previous. Triss What is E(0.120)? Post Discussion らSend Feedback 4 6 tab caps lock
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The ele ic Held Due to point chare at distancex ar^e u s positixc - 0% If is he gativeE2 E E, E2 Ei 2 K.외 2 al +0. (o·リ2 (0.2) 2 .9 2.20 X o-S 328 X lo-4- (o.212 ECo.osm) = (o+005)005) S.28 X to-4- (o-15)1E. 4- 8,20 x 10 3,23 X10 o.12-о O.12 \ 2 o.02)2

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