## 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???

1. Fanduekisses

the radii of both A and B are simultaneously doubled -----> does that mean the force of attraction decreases by 1/16????

2. Fanduekisses

@dan815

3. Fanduekisses

@whpalmer4

4. anonymous

is this chemistry?

5. Fanduekisses

yes, hehe.

6. anonymous

this should be posted in chemistry section then..but anyways, how do you know it is 1/16, were you given a formula?

7. Fanduekisses

$\frac{ (q_{1} \times q_{2}) }{ d^2}$

8. Fanduekisses

distance increased so force decreased.

9. Fanduekisses

2 times 2 = 4 then 4 squared 16... so it decreases by 1/16?

10. 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?

11. 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

12. Fanduekisses

|dw:1444273535708:dw|

13. Fanduekisses

14. 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 -.-

15. Fanduekisses

hehe np, so then it will decrease by 1/16?

16. 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.