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Force between Two Conducting Spheres

  1. Jan 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Two tiny conducting spheres are identical and carry charges of -19.4 μC and +49.3 μC. They are separated by a distance of 2.55 cm. What is the magnitude of the force that each sphere experiences?

    2. Relevant equations
    Coulombs Law: F=kelq1q2l/r2
    Ke = 8.99E9

    3. The attempt at a solution
    The way I write down the numbers is how I put them into my calculator.
    F=((8.99E9)(-19.4E-6)(49.3E-6)) / ((2.55E-2)2)
    F= -8.5982158 / 6.5025E-4
    F=-13222.9385
    but magnitude cant be negative so it would be
    F= 13222.9385 N

    I am not sure what I am doing wrong. I have tried multiple tries at reworking/entering the problem into the calculator to see if I'm somehow messing it up, however I keep getting the same answer.
    F=[((-19.4E-6)(49.3E-6)) / ((2.55E-2)2)] * (8.99E9)= 1.47084967E-6 * (8.99E9) = 13222.9385 N
    F=(8.99E9)(-19.4E-6)(49.3E-6) = -8.5982158 ; (2.55E-2)2 =6.5025E.4 ; F= -8.5982158 / 6.5025E.4 = 13222.9385 N


    Help would be greatly appriciated!
     
  2. jcsd
  3. Jan 17, 2014 #2

    TSny

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    Hello, EmilyfromOH. Welcome to PF.

    Note the absolute value signs in the formula. When you want the magnitude of the force, use the absolute values of the charges.
     
  4. Jan 17, 2014 #3
    When I did the calculations with the abs value, the number tuned out the same.
     
  5. Jan 17, 2014 #4

    TSny

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    Note |(-2)(+3)| = |-6| = 6 (a positive result).

    Likewise |(-2)(-3)| = |6| = 6 (a positive result again)

    So, no matter what the sign of the charges, |q1q2| will yield a positive number. All of the other numbers in Coulomb's law are also positive. So, F must come out positive.

    Realizing this, there is no need to include any negative signs with the charges when plugging into your calculator while calculating F. Just plug in their absolute values.

    (But be careful: There are formulas for other quantities in electricity where you will need to include the signs of the charges and the answers for those quantities can be negative.)
     
  6. Jan 24, 2014 #5
    Ok, thanks a lot, that was very helpful!
     
  7. Jan 24, 2014 #6

    lightgrav

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    wtf? Force is a vector, so it has direction ... a positive direction is NOT the same as a negative direction.
    Your calculator entry is flawless, the correct answer IS -13.22 kN away from the other charge ...
    "negative away" means "toward" the other charge; each attracts the other.
    This is very different than a +19μC charge and a +24μC charge: they would repel, not attract.

    probably you're entering the numerical value wrong into the on-line text-box, or messing the units entry.
     
  8. Jan 24, 2014 #7

    TSny

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    Hi, lightgrav. The question asks for the magnitude of the force on each sphere. So, a positive answer would be required.
     
  9. Jan 24, 2014 #8

    lightgrav

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    the OP knew that in the very first post; but thought that her computed positive value was wrong anyway.
    (probably because it didn't match the mistaken "answer key" - that was her real issue!)
    Your comments (and her textbook) could mislead her into thinking that Coulomb Forces are always positive.
    Sorry, but that approach ignores the most important feature of a Force, so we should carefully avoid it.
     
  10. Jan 24, 2014 #9

    TSny

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    Yes, the OP knew she had to find the magnitude. I thought maybe you had overlooked that point, since you said the the correct answer IS negative. I thought by "answer", you were referring to the answer to the OP question since you suggested that she was not entering the answer correctly into the online text-box.

    Sorry for any misinterpretation.
     
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