Question

Imagine you put all positive charges from a cubic decimeter of iron on the moon, and the negative charges remain on the earth. How big would the electrical force between the moon and the earth be then?

Answer 1

I know im gonna use coloumbs law and have to find how many protons and elektrons a cubic decimeter of iron contains.. But how to do it.. thats h
ard

Answer 2

we have to compute how many atoms of iron there are into 1 dm^3 of iron

Answer 3

Now, what is the density of iron?

Answer 4

7874 kg/m^3

Answer 5

ok! Better is 7.874 g/cm^3

Answer 6

so waht is the mass contained into 1 dm^3 of iron?
keep in mind that:
1 dm^3= 1000 cm^3

Answer 7

oops... so what*

Answer 8

protons and elektrons?

Answer 9

no, the mass of iron into 1 dm^3:
mass = density * volume

Answer 10

mass= 7.874* 1000=...?

Answer 11

since:
1 dm^3 = 1000 cm^3

Answer 12

mass = 7,874 grams
is it ok?

Answer 13

oh yeah 7874 kg?

Answer 14

no, 7874 grams, since the density is 7.874 grams/cm^3

Answer 15

oh sorry yes ofcourse

Answer 16

ok! Now we know that the molecular mass of iron is 55.85. Is it ok?

Answer 17

please take a look to your periodic table of elements

Answer 18

yes thats correct!

Answer 19

ok! So we can say that into 1 dm^3 of iron, there are:
\[\frac{{7874}}{{55.85}} = 140.98\] moles of iron

Answer 20

since 1 mole of iron is 55.85 grams of iron. Is it ok?

Answer 21

yes!!

Answer 22

ok!

Answer 23

Now, 1 mole of iron contains 6.022*10^23 atoms of iron. so
there are
\[140.98 \times 6.023 \times {10^{23}} = 849.12 \times {10^{23}} = 8.5 \times {10^{25}}\]
atoms of iron into 1 dm^3 of iron

Answer 24

\[\Large 140.98 \times 6.023 \times {10^{23}} = 849.12 \times {10^{23}} = 8.5 \times {10^{25}}\] atoms of iron

Answer 25

yes

Answer 26

ok!

Answer 27

Now, we have to keep in mind that the atomic number of iron is 26

Answer 28

so, the number of protons contained into 1 dm^3 of iron is:
\[\Large 8.5 \times {10^{25}} \times 26 = 221 \times {10^{25}} = 2.21 \times {10^{27}}\]

Answer 29

im with you

Answer 30

ok!

Answer 31

Now, how many coulombs are \[\Large 2.21 \times {10^{27}}\] protons

Answer 32

2.21x10^27*1.6x10^-27?

Answer 33

I think:
\[\Large 2.21 \times {10^{27}} \times 1.6 \times {10^{ - 19}} = ...?\]

Answer 34

yes ^-19.....

Answer 35

3.536C?

Answer 36

ok!:
\[\Large 2.21 \times {10^{27}} \times 1.6 \times {10^{ - 19}} = 3.536 \times {10^8}Coulombs\]

Answer 37

yea, and that is q_1? and q_1=q_2?

Answer 38

\[F=k*\frac{ q^2 }{ r^2 }\]

Answer 39

Now, we have
\[\large 3.536 \times {10^8}Coulombs\] on the moon, and:
\[\large 3.536 \times {10^8}Coulombs\] on the earth

Answer 40

what is the mean distance moon-earth?

Answer 41

3,844x10^5m

Answer 42

ok!

Answer 43

the magnitude F of the requested force, is:
\[\Large F = K\frac{{{Q_1}\;{Q_2}}}{{{d^2}}} = 9 \times {10^9} \times \frac{{{{\left( {3.536 \times {{10}^8}} \right)}^2}}}{{{{\left( {3.84 \times {{10}^8}} \right)}^2}}} = ...?\]

Answer 44

7631406250 N

Answer 45

better is:
\[\Large F = 7.63 \times {10^9}Newtons\]

Answer 46

please note that:
\[{Q_1}\; = - {Q_2}\]
so the requested force is an attractive force

Answer 47

oh okey, thanks again! michele, your a saviour!!!

Answer 48

the

Answer 49

thanks! :)

Answer 50

there were alot of chemistry in this one? in the beginng?

Answer 51

yes!

Answer 52

thats why it was so hard!!

Answer 53

thanks alot once again! now Ive to go to sleep
cya

Answer 54

thanks again! :)