## Question

###### This is all 1 problem! Ive been trying this for 4 hours and I just cant...

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

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

Lab 1: Electric Charge, Electric Field and Electric potential In this lab you will use the Charges and Fields PhET lab to study the electric field and electric potential in the space surrounding one or more point charges. 1. The Electric Field and Electric Potential Created by a Dipole Click on the "Grid button. Pick the+InC and the -InC point charges and place them on the grid 5 m away from each other. Verify the distance with the measuring tape. This is how your computer screen should look like: PLETE Consider a Cartesian xy coordinate system with the origin fixed on the positive point charge, the x axis horizontal the y axis vertical. A. Click on the "Electric Field" button: the electric field vector will be displayed on the screen. You can use the electric field sensor to measure the magnitude and direction of the electric field at different locations. B. Calculate the magnitude and direction of the electric field at the following locations, then compare with values obtain from the simulation: Emagnitude Edirection magnitude Edition calculated Calculated Moure Mead" 1 Point A located midway between the changes 2 Point Blocated at m. -1.5m) 3 Point located at ( xm, y1.5m) With your phone take a picture of your calculations and insert it here to show your work C. Calculate the electric potential at the same locations and compare with the values "measured with the equipotential meter. Do not forget to include sign. Point MOV

C. Calculate the electric potential at the same locations and compare with the values "measured with the equipotential meter. Do not forget to include sign. calculated 1 Point A located midway between the charges 2Point Blocated at my1.5m) 3Point located at -4m, 1.5m With your phone take a picture of your calculations and insert it here to show your work D. Click on the pencil drawing too attached to the equipotential meter. you will see a green equipotential line. Use the tool to draw equipotential lines every 0.5 m on each side from the midpoint. Take a screen shot of the picture you got and copy it here. E. Calculate the change in the potential energy AP of a test charge 4. - I pc when moved from point Pxlm, y 0) to point Of 2.3 m, y-0). IpC=10c Record your value: AP 2. The Electric Field and Electric Potential Created by a Three Point Charge Distribution You will study the electric field and electric potential created by three point charges at following locations: +InC at (x-O and y +1.5 m), +InC at (x=0 and y=-1.5 m) and -InC at (x - 5 m and y = 0), as shown in the picture: A. Calculate the electric field at x = 25 m and y = 0 and compare with values obtain from the simulation: Point E made m e direction calculated Calculate "Ma d Measured Point A located at 3 -09 With your phone take a picture of your calculations and insert it here to show your work B. Draw equipotential lines at this point. Draw equipotential lines by e moving the n cach side of this

2. The Electric Field and Electric Potential Created by a Three Point Charge Distribution You will study the electric field and electric potential created by three point charges at following locations: +InC at (x = 0 and y = +1.5 m), +1nC at (x = 0 and y=-1.5 m) and -InC at (x = 5 m and y = 0), as shown in the picture: Od Charge And Felds PRETE A. Calculate the electric field at x = 2.5 m and y = 0 and compare with values obtain from the simulation: Point E magnitude Edirection E magnitude E direction calculated Calculated "Measured" "Measured Point A located at (x2.5m, y0) With your phone take a picture of your calculations and insert it here to show your work. B. Draw equipotential lines at this point. Draw equipotential lines by moving the equipotential meter 0.5 m away on each side of this point. Take a screen shot of your results showing the electric field vectors at different locations and the equipotential lines that you have drawn. Copy the picture here.

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

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 figure). Find the magnitude and direction of the electric field at point, which has coordinates (0, 0.400) m EEN: E,/ - (9*10° N-m’IC) 0.400 m E. - 3.93x10' NC E.. - E.cos(90')- 0 E,-E, sin(906) 3.93x10' N/C E, - ,-0.500m E, 1= (x10* N.mC") 6.500 m E , -1.80x10NIC Es. - Ecose - 1.08x10N/C E,-E, sin 0 - -1.44x10' N/C E.-E.E-1.08*10'N/ C E ,-E,, +E, -2.49x10'N/C E - 2.71x10'N/C E-VE+E; 6- tan 66.6 ious

## Answers

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 - kla, 1 here r = PA= 5=205m 2,2 k=9x109 Nm²/C2 TE l = 19x109) (1x109) N 1 205)2 ł TE il = 1.44 N/C In direction pho ArteCosol 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 a El = k/a, 1 r = All = 5=2,5m r, 2 TE2l = 98109 x | 1x1591 (2.52 11 =1.44NIC (Ex1 = Eg (Oso 16u1 - E₂ =1.44NIC LE2y1 = Esino = EO Lrence L Net Electric field at A (EL Ex Ext Esa Ex=1.44 +1.44 Ex = 2.88 NIC y = Eigt Ezy Ey = o to Ey = I - Ez 6,1 = {(2,399 484 - 2.8 KMtanoz 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=PD+ 8 Sitil-0-6207ी गी + ) E alongp to B पित Ex - E, cono दापणे E - (0- 6 207 ) (०.११) IS = J/५ Eu 0.5705 w/C 3. या ___(009 - EN (00= 3.1 Ey = Ei since E = (.62070/07 TEL02445 MIO H5.5 (3-1oy) उन्ट- 10 30-34 पण130²= 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 due to a LEI - klaal Inable EL = (9X169) |- XIE (4 ) E2 - 2 NIC de 2 = 4.5m) E - E (and and Buzhur Ex=2 x 0.207 Blo=2.1213m In = 1.414 NI kus 0 = 165" (100 t - - Esind -ve because 4 =10st G. 1213 - --2X0-707 sind = + = 0.707 Hence E = Ext Ex & Cond = + = 0.707 = 0 Sost 1.414 = = 1.9845 Nc Ey = Eigt Ezy Ey = 0.2445+ (-1.914) Ey = - 1.1695 Nic / E = Ety = $11.98457(-1.16952 E = 2.3035 N/C - To Angle 6400 of E with 0 - tal-17695 It anyt sar451 0 -30-SPoint 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, (050 Heater = kan 700) LEQ = 0.2022 * 0.9744 0 = ta plot TEui - 0.19 707 Nic o = 12.995° Fly = E, sino (5o=0.9744 Ein = 0.2022 x 0.2249 sino = 0.2249 Eig=0.0455 NIC r = Pc ² = po²+ CD? 1 r. -16.5 1 1 (axto) 141041 roce (2661054) - ( Tush stance Era 2 Mic along to = fant (1 ) Ema-E cond $ = 45 EL =-2 (0.707) Cor - - Ex= -1.414N16) Sind = + 2 = 0.767 Egy = - Eg Sind & I-10 2 lay = -2(0.707 2 = 0 = 0 0² +DC2 y = -1.414 NT r = 1.52 +152 2-4.57 (= 2.1234 rez606708ul OD X En I 0-7c7t = 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+ (-0.86735 E = 1.2186 N(C) tot Eu te - 0.06434 ) - 1.2 169 - lo = 3.0 20°) Now electric potential v= * a ATA V = V + 2 Vi= Potential - dues duetoh t = avro x laco" +(4xto?) Elka V - 4x109x1809 - 9x109 x 18109 2 . 5 2.5 610 x 18109 2 ) IV - o At B - V = V tu. - Ka thre 1v (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 ) 1X109) - 4.49087- .7273) tel. 3492 -4.2427 1V= -2.8935v) Equipotential I because at midpoint L A . Electric field tine oil, is along x anis P 0.5 then Equipotentialline 257 25 kas must be along yanis because Electric field ý always t to Equipotential line. (E) OP = Upinat gust tax so U = Electrostatic Co har Potential enerry or -E Unitial = R +U₂ twihat k9,9 + 1929. Initial - i Visite =lawo(1x)(1+2) . . ((3) چکاد +(9x19) (-1810") (10th) 9. = 1pc (105) (90= 15120) Vinitial = -3.429x16"J) Similarly Ufinal = U, tu2 - Ufinal = hango tha az $ 12 Ufinal= (9x19) (1x109) (10 ) (4.8) - + 198109) (-1910) (1672) (002) - - 430125 X 1037 BP = final - Vinitial 3p = -43.125 X 10 -(~3-4296103) TOP = – 39.696 X 13 J Vam 352,21% = 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 to = fon (les Eix = E(GO 0=30.964. Ex=(1.0588) (0.8575) Co2 0.8585 Nle. TEm = 0.9079 wle) sino=o.sius we F E Sihe AD² = 2,² = 1.5 +2.52 I /10188 (0.3145) ² = 825 m T -0.5448 NLC) 21 = 2-9155 m Now because a, and an are taurt and Symmetrically Perced from D velr, at Hence (Eil=fal=100588 wc "Ex = 0.9079010 [ Eu = Ein] and - - Eig . Ezy is in - - (-0.1421 8 opposite direction to - 0.6148 ste Eig] Tty also 256 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 SSB NI Ey - Eryt by t Ezy - -0.5448 +0.3448 to Ey = 0 the = tem xe = toimus) = 0 deten - 3.2550 2to2 [E = 3.2858 wl• because Electric al Fanpotential field is along nants at D. Hence the Equipotential line must be along y any because Equipotential line is always I to Ravipotenked Electric field.

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