Question

Two charges attract each other with a force of 1.5 N. What will be the force if the distance between them is reduced to one-ninth of its original value?

Answer 1

do you know coulombs law

Answer 2

Yes, (k)(q)(q2)/d^2

Answer 3

what is in question says is distance between them reduced to 1/9 put d=1/9 *d
right and solve

Answer 4

So what would I set the two equations to?

Answer 5

F is proportional to the inverse square of distance.

Answer 6

do u know the answer LOL:)

Answer 7

I get answers but they seem way off...

Answer 8

When I solve them, I have so many variables, so I can set a system of equations

Answer 9

*So i can't set

Answer 10

OK F = 1/d^2 = consider as proportional

Answer 11

now d becomes one by ninth mean =1/ 9 *d right

Answer 12

right

Answer 13

F=1/(1/9)^2 d2

Answer 14

right

Answer 15

right

Answer 16

F =1/(1/81)d2

Answer 17

F=81/d^2 right

Answer 18

yeah, i have that

Answer 19

F/81=1/d2

Answer 20

1.5/81 = 0.0185 N right

Answer 21

Yes

Answer 22

so i get: 0.0185d^2

Answer 23

hey wait if distance ids reduced means force should be larger then initial right but we are getting less force here?????

Answer 24

its like this F2=81/d^2 right

Answer 25

I have that

Answer 26

F2=F1*81
=1.5*81
=121.5 N is right

Answer 27

ohhh. I didnt see that you could set the \[(k \times Q1 \times Q2)\] as a constant