Question

Will someone help me figure this out?

The charges on two metallic balls are 5.0 and 7.0 coulombs respectively. They are kept 1.2 meters apart. What is the force of interaction between the two?
(k = 9.0 * 10^9 newton·meter^2/coulomb^2)

A) 2.2 x 10^11 newtons
B) 4.1 x 10^-7 newtons
C) 5.0 x 10^11 newtons
D) 7.0 x 10^-7 newtons

Answer 1

Have you learned Coulomb's law?

Answer 2

No not yet. I saw the formula but I don't get what it means o-o

Answer 3

F =k q1* q2/r^2
This is it right?

Answer 4

Cool cool - so Coulomb's law is

\[ |\textbf F|=k\frac{q_1q_2}{r^2}\]
which gives the magnitude of the force between two charged objects

q1 = charge of first object
q2 = charge of second object
r^2 = square of the distance between them
k = Coulomb's constant (whose value they give)

Answer 5

yeah, what you wrote is right ^_^

Answer 6

So q1=5.0 and q2=7.0
Multiply that right? And I would get 35.0 then I would divide that by 1.2 and that's 29.1666666667

Answer 7

Then what o_o
I dont get what I do after

Answer 8

(1.2m)^2 = 1.44m^2

Answer 9

Then you multiply 35C^2/1.44m^2 by k

k = 9x10^9 C^2/m^2

Answer 10

What.. Now youre confusing me ._. So I didnt do that right?

Answer 11

not quite - remember it's

\[F=k\frac{q_1q_2}{r^2}\]

you got the
\[q1*q2 = 7C*5C = 35C^2\]
but you divide by r^2, not just r

\[r^2 = (1.2m)^2 = 1.44m^2\]

Answer 12

But where does the C^2 come in?

Answer 13

Wait, what the heck is C? O.o

Answer 14

That's the units on the charges -

5 Coulombs = 5C
7 Coulombs = 7C

and when you multiply units, they acts just like normal parts of the equation so
\[1C * 1C = 1C^2\]
likewise
\[5C*7C=3C^2\]

Answer 15

But 5*7 is 35..

Answer 16

Im sorry Im not really catching it .-.

Answer 17

Sorry, mistype - 35C^2 you're right

Answer 18

So like, as an example when you're finding the area of a rectangle



What would the answer for the area be?

Answer 19

(this rectangle is in no way related to anything in this problem - it's just an example to look at multiplying units)

Answer 20

4x6 = 24 o

Answer 21

yah, but the units on it are

\[A=(4m)(6m) = 24m^2\]
right?

Answer 22

Oh oh okay Im getting this xD

Answer 23

^_^ !

Answer 24

9x10^9 C^2/m^2
So that^
I would just do that equation then? And thats my answer o-o

Answer 25

Coulomb's law is what gives the answer you're looking for

\[F=k\frac{q_1q_2}{r^2}=(9.0\times 10^9 N m^2/C^2)\frac{35C^2}{1.44m^2}\]

See how in the units, C^2 cancels, m^2 cancels, so you're only left with Newtons for force?

Answer 26

35/1.44?

Answer 27

24.30??

Answer 28

I should just divide those to numbers then? or those numbers cancel each other out too?

Answer 29

You divide the numbers, the units are what cancels out

Answer 30

\[F=\left((9\times 10^9)\frac{35}{1.44}\right)\left((N\cancel{m^2}/\cancel{C^2})\frac{\cancel{C^2}}{\cancel{m^2}}\right)\]
\[=\big(2.19\times10^{11}\big)\big(N\big)\]

Answer 31

Oh. So all I had to do was that ._.
Atleast i know how to do it know for next time. Thank you very much.

Answer 32

Sorry to complicate things :/

Answer 33

Oh no no. You helped me a bunch [: You have no idea xD