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:
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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
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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=...
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    Four Equidistant Particles in electric fields

    well that would be 3i+4j, correct? Would I do the cross product?
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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...
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    Four Equidistant Particles in electric fields

    gotcha, I'll do this when I have break between classes. :smile:
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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?
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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?
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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...
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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:
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    Four Equidistant Particles in electric fields

    so is K the positive test charge in the center of the field?
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    Four Equidistant Particles in electric fields

    ohhh, but isn't electric field E=(kq)/r^2?
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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?
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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:
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    A screen for detecting electrons

    ok the numerical value is correct but the units are still wrong! lol
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