Question

I still need a bit oh help :) ..

An attractive force of 7.2 N occurs between two charges that are 0.05 m apart. If one charge is 4.0 μC, what is the other charge?
(Points : 5)
+0.5 μC

+2.0 μC

2.0 μC

4.0 μC

Answer 1

what is the relation of force with charges and distance! i mean the coulombs force

Answer 2

Have no clue >_<"..

Answer 3

oh do you mean coulomb's law?

Answer 4

yeah!

Answer 5

equation is F=K(q1q2/r^2)

Answer 6

yeah

Answer 7

just plug in the values and solve for one of the charges

Answer 8

7.2=(9*10^9)(4*q2/.05^2) thi is the part where I get lost on..
I'm discombobulated on how I would rearrange the problem to find q2...

Answer 9

Kinda. Can you explain what's happening in the picture so I may acquire an even awesomer understanding ? :)

Answer 10

oops this thing is soo messy!

Answer 11

lol it's readable it's just physics and me just don't mix :P

Answer 12

whale where are you finding problem in calculating q2?

Answer 13

when I get my answer it's just 7.2= 14400000000000q2

Answer 14

do I divide that by 7.2 or did I just do th whole thing wrong?

Answer 15

wait!!

Answer 16

lol okay

Answer 17

im getting 5uC

Answer 18

as the answer for q2 or my equation ? ( I got 5uc the first time to but it was not an answer choce so I thought it was wrong)

Answer 19

sorry 0.5 uc

Answer 20

isnt that the answer?

Answer 21

Oh yes :) how did you get it?

Answer 22

i think you didnt notice that it the other charge was 4uC so its gunna be 4*10^-6
everything else is fine. if this dosent get you to the answer then mayb you hav a calculaton error!

Answer 23

:)

Answer 24

Thanl you so much ! ^(^o^)^ you were right I totally missed the 1.0 * 10^-6 C That tis why I was wrong . Thank oce more and you got another medal ^_^

Answer 25

np