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Electrostatic force

  1. Dec 6, 2003 #1
    The electrostatic force between two point charges is 5.7*10^-4N. The charge on one of the point charges triples and the distant between the points doubles. What is the force between the two point charges?
  2. jcsd
  3. Dec 7, 2003 #2
    5.7*10^-4N/2 = 2.85^-4N What have i done wrong?
  4. Dec 7, 2003 #3


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    What is Coulomb's law?
  5. Dec 7, 2003 #4
    Using Coulomb's Law you're given:

    [tex]\vec{F_{e}} = k\frac{qQ}{r^2} \hat{r}

    where [tex]q[/tex] and [tex]Q[/tex] are two different point charges.

    By tripling one of the charges (it doesn't matter which one) and doubling the distance you now have:

    [tex]\vec{F_{e}} = k\frac{3qQ}{(2r)^2}\hat{r}=k\frac{3qQ}{4r^2}\hat{r}[/tex]

    You should now be able to find the force between the two charges by plugging in the initial values of the two point charges and the original distance r.

    It always helps to work directly from the law or formula that's given.
  6. Dec 7, 2003 #5


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    Science Advisor

    What rogue said is exactly right but it may still look confusing. Just look at the constants added to the formula. The original formula was this
    [tex]\vec{F_{e}} = k\frac{qQ}{r^2} \hat{r}[/tex]

    The new formula is this:

    [tex]\vec{F_{e}} = k\frac{3qQ}{4r^2}\hat{r}[/tex]

    The only things that changed are the 3 in the numerator and the 4 in the denominator.
    Just multiply your original force by 3/4 and you should get the answer :)
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