anonymous
  • anonymous
Calculate the force experienced by a positive test charge of 8.1 x 10^-9 coulombs if the electric field strength if 9.4 x 10^7 newtons/coulomb. I know (or at least THINK I know) that the equation needed for this is something like: F = Kq1q2/r^2 BUT, I need someone to explain to me where to plug in the numbers. Assistance, please?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Astrophysics
  • Astrophysics
Ok, so when you're doing physics, it's best to list out all the things we are given, it will make your life 100x easier! So lets go through the question and list what we have and what exactly we are looking for, start doing this for all your problems! So we are looking for the electric force, \[\vec F_e =~ ?\] \[q_{test} = 8.1 \times 10^{-9} C\] \[\vec E = 9.4 \times 10^7 N/C\]
Astrophysics
  • Astrophysics
This question is similar to your previous but we have a positive charge meaning the electric force and field are in the same direction, hence we use the formula we did previously. \[\vec E = \frac{ \vec F_e }{ q_{test} }\]
anonymous
  • anonymous
PFFFT can you tell I'm not a physics major? :B Okay, so if we are to use the same equation as before, it'll look like this again: F = 8.1 x 10^-9 + 9.4 x 10^7 right?

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Astrophysics
  • Astrophysics
Well you don't have to be a major in physics, this is useful for all sciences! Not quite, if you have trouble with the algebra, look at the units and maybe that will help you. \[\vec E = \frac{ \vec F_e }{ q_{test} } \implies \vec F_e = \vec E \times q_{test}\] notice I am multiplying it and not adding it!
Astrophysics
  • Astrophysics
Does this make sense? |dw:1434944865402:dw| what you do to one side you must also do to the other.
anonymous
  • anonymous
Hm...alright, I think I got it F = 8.1 x 10^-9 * 9.4 x 10^7 = 0.7614 = 7.6 x 10^-1 ?
Astrophysics
  • Astrophysics
Looks good :)
anonymous
  • anonymous
Whoo! Thanks again. :3
Astrophysics
  • Astrophysics
Np :)

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