In a rectangular coordinate system a positive pointcharge q = 6.00 - 10 C is placed at the point a= +0.150 m, y = 0, and an identical point charge isplaced at r = -0.150 m, y = 0.%3DPart DFind the r- and y-components, the magnitude, and the direction of the electric field at the a0.300 m, y = 0.Express your answer in newtons per coulomb.Πνα ΑΣΦEx =N/CPart FExpress your answer in newtons per coulomb.1OLE%3DN/CPart GExpress your answer in degrees.ΑΣΦcounterclockwise from theta direction12

Answered: Image /qna-images/answer/da7b0c78-88df-4057-8b73-bca0b31a7160.jpg

Answered: In a rectangular coordinate system a…

Similar questions

Answer all parts of the PHYSICS problem please
H
I Rew A spherically symmetric charge distribution produces the electric field E =( 5700 r2)r N/C, where r is in m. Part A What is the electric field strength at r = 19.0 cm ? ? N/C Submit Request Answer Part B What is the electric flux through a 38.0-cm-diameter spherical surface that is concentric with the charge distribution? Nm2/C Submit Request Answer Part C How much charge is inside this 38.0-cm-diameter spherical surface? C
Needs Complete typed solution with 100 % accuracy.
Question: In the figure particle 1 of charge q1 = -4.99g and particle 2 of ... Continue to post In the figure particle 1 of charge g= 4.99q and particle 2 of charge g2 = +3.39g are fixed to an x axis. As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero? 19 questions remaining g am t . nyda ma My Textbook Solutions Show transcribed image text View comments (1) > Organic. University. Introductio.. 4th Edition 2nd Edition 4th Edition Expert Answer View all solutions Anonymous answered this Was this answer helpful? 2,147 answers Create an As ter the qwen inviting air. Adjust the mood in enyizeroom L-X Sho sam's Elec rie fuld olu to : - K4 99 q Elac tue fuld dut -K 3-39 to EL - Ka2 tet u Pu the pount where Egut =D- asseume So, - 4 99 Kg 3.99 kg 3.99 4 99 (L-x)! Please explin neatly and Cleally how to get from e step by step, to finally Solve foe X. 3-99 x 4 99 CL-x)L E5.6892 L Ans Cornment > Thank ypu! Practice with similar questions…
Electric Fields 1. A wire has been given a net positive electric charge Q evenly distributed along its length. The wire has a length of L and is located on the x-axis with its left end at the origin as drawn. Set up the integral you need to do to find the electric field at a distance d to the left of the origin. Some steps you may wish to follow are: a. Figure out the direction the electric field will have at our location x = -d. This will be the same as f b. Write the differential charge dą in terms of the charge per length and a differential length. c. ris a function of x which means we have to integrate to find the electric field. Write down r. d. Put everything together (including the vector direction) into one integral that would allow you to find the E-field. Write this in the box below. e. You are not required to complete the integral and no credit is given for completing the integral. d dE =
need help on part A, C, D
I do not know how to solve the attached phyiscs question.
y y :P2 B В Figure 2 Figure 1 a. i. Point P1, the intersection of the three bisectors, locates the geometric center of the triangle and is one point where the electric field is zero. On Figure 1 above, draw the electric field vectors EA, EB, and Ec at P1, due to each of the three charges. Be sure your arrows are drawn to reflect the relative magnitude of the fields.
P8