Fanduekisses
  • Fanduekisses
********** help pls. ;'( IS THIS THE RIGHT ANSWER? How would you expect the force of attraction between positive ion A and negative ion B be affected by the following changes???
Mathematics
chestercat
  • chestercat
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Fanduekisses
  • Fanduekisses
the radii of both A and B are simultaneously doubled -----> does that mean the force of attraction decreases by 1/16????
Fanduekisses
  • Fanduekisses
Fanduekisses
  • Fanduekisses

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anonymous
  • anonymous
is this chemistry?
Fanduekisses
  • Fanduekisses
yes, hehe.
anonymous
  • anonymous
this should be posted in chemistry section then..but anyways, how do you know it is 1/16, were you given a formula?
Fanduekisses
  • Fanduekisses
\[\frac{ (q_{1} \times q_{2}) }{ d^2}\]
Fanduekisses
  • Fanduekisses
distance increased so force decreased.
Fanduekisses
  • Fanduekisses
2 times 2 = 4 then 4 squared 16... so it decreases by 1/16?
anonymous
  • anonymous
F = k q1 q2 / d^2 exactly what i am thinking :P If you double the distance BETWEEN those two atoms, wouldn't be 1/4?
Fanduekisses
  • Fanduekisses
Why does it say "the radii of both A and B are simultaneously doubled" ? I thought it meant like the radii of both so 2 * 2
Fanduekisses
  • Fanduekisses
|dw:1444273535708:dw|
Fanduekisses
  • Fanduekisses
wait radii is the plural of radius? lol xD
anonymous
  • anonymous
oh it says "radii of BOTH" so i think you are right then |dw:1444273677199:dw| sorry i keep missing important parts of the question -.-
Fanduekisses
  • Fanduekisses
hehe np, so then it will decrease by 1/16?
whpalmer4
  • whpalmer4
let's say the two ions both have a radius of \(s\). the center-to-center distance is therefore \(s+s=2s\). if we use the usual formula \[F=\frac{kq_1q_2}{r^2}\] (assuming they can be treated as if they are point masses) then we have a force of \[F_{orig}=\frac{k q_1 q_2}{(2s)^2} =\frac{k q_1 q_2}{4s^2} = \frac{K}{4s^2} \] (K represents all of the quantities which will not change) Now we double the radius of each so the center-to-center distance becomes \(2s+2s=4s\), and our equation becomes \[F_{new}=\frac{k q_1 q_2}{(2s+2s)^2} = \frac{K}{(4s)^2} = \frac{K}{16s^2}\] \[\frac{F_{new}}{F_{orig}} = \frac{\frac{K}{16s^2}}{\frac{K}{4s^2}} =\frac{K}{16s^2}*\frac{4s^2}{K} = \frac{\cancel{K}}{16\cancel{s^2}}*\frac{4\cancel{s^2}}{\cancel{K}} =\frac{4}{16}=\frac{1}{4}\] So the new force is 1/4 of the original force.

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