Electrostatic force question with coulombs law

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SUMMARY

The discussion centers on calculating the electrostatic force between two charged particles using Coulomb's Law. The first particle has a charge of +4.20 x 10^-6 C and is 12.0 cm away from a second particle with a charge of -1.70 x 10^-6 C. The correct application of Coulomb's Law, F = (K*Q1*Q2) / R^2, where K = 8.99 x 10^9 N m²/C², yields a magnitude of 4.457 N. The user initially misinterpreted the sign of the force, mistakenly believing it should be negative due to the attraction between the charges.

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  • Basic algebra for manipulating equations
  • Conversion of distance units from centimeters to meters
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Homework Statement



A particle of charge of +4.20 x 10^-6 C is 12.0 cm distant from a second particle of charge of -1.70 x 10^-6 C. Calculate the magnitude of the electrostatic force between the particles.

Homework Equations



Coulombs law is F = (K*Q1*Q2) / R^2 where k = 8.99*10^9, q1 and q2 are the charges, and R is the distance in meters

The Attempt at a Solution



Im hoping I am just making a simple mistake and not seeing it for this question cause this really seems like plug and chug.

K Q1 Q2 R
so I did (8.99 x 10^9)(4.20x10^-6)(-1.70x10^-6) / (.12)^2

= -4.457 N

that gave me a wrong answer so i tried different variations of putting the decimal like -4457.541, 44.57 and -44.57. What am i doing wrong, i might have just fudged up since I'm doing this by hand but I'm not seeing it. Either that or I have done something fundamentally wrong and don't know any better. It's been a while since I've done a class like this so any help is appreciated.
 
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The problem asks the magnitude of the force, which is positive. The number itself is good.

ehild
 
heh well like i said it's been awhile, I kept thinking since they attracted the force would be negative. Thanks a lot though, guess i just needed a second pair of eyes
 

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