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Fanduekisses
 one year ago
********** 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???
Fanduekisses
 one year ago
********** 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???

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Fanduekisses
 one year ago
Best ResponseYou've already chosen the best response.1the radii of both A and B are simultaneously doubled > does that mean the force of attraction decreases by 1/16????

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this should be posted in chemistry section then..but anyways, how do you know it is 1/16, were you given a formula?

Fanduekisses
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ (q_{1} \times q_{2}) }{ d^2}\]

Fanduekisses
 one year ago
Best ResponseYou've already chosen the best response.1distance increased so force decreased.

Fanduekisses
 one year ago
Best ResponseYou've already chosen the best response.12 times 2 = 4 then 4 squared 16... so it decreases by 1/16?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0F = 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
 one year ago
Best ResponseYou've already chosen the best response.1Why 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444273535708:dw

Fanduekisses
 one year ago
Best ResponseYou've already chosen the best response.1wait radii is the plural of radius? lol xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh 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
 one year ago
Best ResponseYou've already chosen the best response.1hehe np, so then it will decrease by 1/16?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0let's say the two ions both have a radius of \(s\). the centertocenter 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 centertocenter 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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