anonymous
  • anonymous
Can someone please check my work on this equation so far? Here is the question... Two point charges each carrying a charge of + 4.5 E – 6 C are located 4.5 meters away from each other. How strong is the electrostatic force between the two points and is this force a repulsive force or an attractive force (k = 9.0 E9 Nm2/C2)? So the values I have are as follows... F = ? k = 9.0 E9 Nm^2/C^2 q1 = 4.5 E - 6 C q2 = 4.5 E - 6 C r = 4.5 m Am I on the right track? Thanks!
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ash2326
  • ash2326
@SarahCatherinez yes you have got the values of each of the parameter correct. Seem to be on the right track. Go on and plug in the values in the formula for force.
anonymous
  • anonymous
Great! I'll go ahead and plug in the values, if you wouldn't mind checking my work from there.
ash2326
  • ash2326
yes, no problem

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anonymous
  • anonymous
F = (9.0 E9) * (4.5 E -6) * (4.5 E -6)/(4.5^2) F = (9.0 E9) * (4.5 E -6) * (4.5 E -6)/(20.25) F = (9.0 E9) * (4.5 E -6) * (2.2 E -7) F = .00891 N
ash2326
  • ash2326
There seems to be a mistake in 3rd step.Could you check?
anonymous
  • anonymous
Hm, that's odd. I just entered it into my calculator and got the same answer.
ash2326
  • ash2326
yes, sorry. Your answer is close to the correct. I would show you another way you have this \[F=\frac{9*10^9*(4.5 *10^{-6})^2}{4.5^2}\] \[F=\frac{9*10^9*(4.5)^2 *(10^{-6})^2}{4.5^2}\] \[F=\frac{9*10^9*\cancel{(4.5)^2} *(10^{-6})^2}{\cancel{4.5^2}}\] \[F=9*10^9*10^{-12} N\] Do you follow this method?
anonymous
  • anonymous
Oh yes, that makes sense.
ash2326
  • ash2326
I think you can find the answer now, can't you?
anonymous
  • anonymous
I believe so. Thank you so much for your help! :)
ash2326
  • ash2326
No problem

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