physicss

Two point charges experience an attractive force of 5N when they are separated by 1m.what force do they experience when their separation is 2m? What is the value of the charges
Using coluomb's law
F=KQ1Q2/r^2
5=8.99*10^9*Q1*Q2/1

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asked by David
  1. force inversely proportional to distance squared

    (1/4)5

    You can only do the second part if you know that the charges are equal.
    then as you pointed out
    Q^2 = 5/k

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    posted by Damon
  2. Pls be mathematical

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    posted by david
  3. I can not be more mathematical than to say if you double the distance you divide the force by 4

    2^2 = 4

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    posted by Damon

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