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this is all 1 problem! Ive been trying this for 4 hours and I just cant figure all these answers out. please circle all answers and remember the box with values, thank you so much and Ill be sure to give you a good rating and Ill also Venmo/Cashapp you 20 for this if you want. thank you
Lab 1: Electric Charge, Electric Field and Electric potential In this lab you will use the Charges and Fields PhET lab to stu
C. Calculate the electric potential at the same locations and compare with the values measured with the equipotential meter.
2. The Electric Field and Electric Potential Created by a Three Point Charge Distribution You will study the electric field a
QQo 180.0 des 3951 Equipotential -373.6 V x 1 nC -1 nc Sensors 1 meter
0.0 deg. 3424 Vin Equipotential 1546 V 0.8 cm +1 nC YASIC

here is the example problem to help lead you in the right direction

Lab 1 Example Problem Charge: 7.00 C is at the one, and charge : -5.00 C is on the x-axis, 0.300 m from the origin (see figur
0 0
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Given. Charse at P a = tinc = 1x09 Charse at a q = - Inc = -180 C tina tinc 2 Inc? 5m Now Electric Field at A due to a El -Cosol Ex= E (oso (Eal = E,(1) = 1.440) TEyl - E, Sino Figl = É, (o) LEyl=0 [Sino zo] Similarly Electric field at A due to atanoz otan (te) - 0=thi 10 1020 At point B B(15) ALLD Electric field at B due to a Eix 11/2011 हा-479) | 1 1. 51 In LABD P=PD130²= Bp²+pQ² B6²=(1-5 +0.5² Bo² = 4.5m L 10 t - - Esime lin-y direction - uco - - 1-414 NIC Simlarly Electric field at B duePoint Similarly, At JE, I = k19,1 Tuinch ET Cat Ey hi 2 Tu A ! 4PC) =(9x109)/18109 (44.5) Ei= 0.2022 NIC alory ptot Ex = E, (t = Ent Ex = 0.19 70 7-lary E = -1.21 69 ale - Eig tEgy = 0.oros-1.414 - Ey = -0.06434 Nic ons E = JE 21E2 - J-16 2169) 2+ (-v (9x109) (1804) (9x109 ) 18169 3.808 ) (2.12 13) 2.3634 - 4.2427 V Potential at v= vit ₂ v=halth V- axto) (1x) (9x109 ) 1X10E Unitial = R +U₂ twihat k9,9 + 1929. Initial - i Visite =lawo(1x)(1+2) . . ((3) چکاد +(9x19) (-1810) (10th) 9. = 1pc (105)2,21% = 1Xc6% ، ب۱۸ = ۱۰ عيه az --inee -19/9 TE=klail HS seega 2 .5 Ei=9x109 x (18109) a tots) In A AED Teil = 1.0588 Nic to6 2 9X/09 x Al xro / - 6 = Ex- E En = Es 1-44NC (oso - E₂11 1.44 sino = Ez (0) E- Eixt Ent Ex - 0.9079+0.9019 +1.44 G - 3.2• because Electric al Fanpotential field is along nants at D. Hence the Equipotential line must be along y any because Equipo

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