Recent content by C6ZR1
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Four Equidistant Particles in electric fields
ohhhhhhhhhhhh ok. This makes sense now. lol Thanks for your time and help. Physics is starting to make more sense now :smile:- C6ZR1
- Post #38
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
the thing is for the answer it says Number: ____ N/C or V/m and I cannot use vector letters, that's what is confusing me- C6ZR1
- Post #36
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
ok I think I'm completely lost. :confused: I find the electric field from each charge: E1= (8.99E9*1.602E-19*7.59E-9) /(0.065)^2= 2.58724 E-15 C E2= (8.99E9*1.602E-19*-1.09E-8) /(0.065)^2= -3.71554 E-15 C E3= (8.99E9*1.602E-19*1.15E8) /(0.065)^2= 39.2007 C E4=...- C6ZR1
- Post #34
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
well that would be 3i+4j, correct? Would I do the cross product?- C6ZR1
- Post #32
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
ok so I found that with the formula F=kq1q2/r^2 in the diagonal direction where the charges are being repulsed that F= 6.50731 E-15 and diagonal direction where charge are drawn inward F=-37.1554, if my calculations are correct, would I just take the dot product and multiply them together to get...- C6ZR1
- Post #30
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
gotcha, I'll do this when I have break between classes. :smile:- C6ZR1
- Post #29
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
ohhhhhhhhhh, then in that case would I Just find F for each particle then add them up to get the sum?- C6ZR1
- Post #27
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
ohh wait, is it F=kq1q2/r^2 where q1 is one of the particles and q2 is the fixed positive test charge?- C6ZR1
- Post #25
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
that makes sense, but if we have a constant Q in this case, the positive test charge, as well as equidistant particles from that test charge then wouldn't E be the same for each particle? ie. E1= 8.99E9*1.602E-19/ (0.045962)^2 Im kinda confused on how they want use to find the magnitude of the...- C6ZR1
- Post #23
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
omg, not k, I meant q. I have no idea why I typed k. :redface:- C6ZR1
- Post #21
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
so is K the positive test charge in the center of the field?- C6ZR1
- Post #19
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
ohhh, but isn't electric field E=(kq)/r^2?- C6ZR1
- Post #17
- Forum: Introductory Physics Homework Help
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Four Equidistant Particles in electric fields
lol, well good morning. So if the vector fields are electric fields produced by each particle then was my original attempt correct my finding the electric fields and adding them up?- C6ZR1
- Post #15
- Forum: Introductory Physics Homework Help
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A screen for detecting electrons
* face palm * I need to stop doing homework so late, I got it now. Thank you soooo much for your help and time. I really appreciate it! This will certainly be a question I'll review for an upcoming exam :smile:- C6ZR1
- Post #34
- Forum: Introductory Physics Homework Help
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A screen for detecting electrons
ok the numerical value is correct but the units are still wrong! lol- C6ZR1
- Post #32
- Forum: Introductory Physics Homework Help