Question

The electric force between two charged objects is 0.24 newtons. If the distance between them is increased by a factor of 4, what will be the new force?
Select one of the options below as your answer:

A.
0.03 newtons
B.
0.06 newtons
C.
0.12 newtons
D.
0.36 newtons
E.
0.96 newtons

Answer 1

would this be b

Answer 2

if it is increased by 2 then its b

Answer 3

so its c?

Answer 4

sorry for asking so many questions im just stressing out

Answer 5

choose b i am not sure..

Answer 6

but if u just said its b if it increases by two then wouldnt it be c or a?

Answer 7

okay choose a

Answer 8

If we're not considering anything about the objects, then we'll just consider the charges to be "point charges." That means that, for the sake of having an idea in our minds, we'll say the charge is at one point and the electric field comes from that point.

So that's what we're thinking about: point charges.

The formula for the force due to static electric charges is this:\[F=k\frac{q_1\ q_2}{r^2}\]I double checked my formula at this link. It's a good site!

Answer 9

So we want to consider what happens when only the \(\boldsymbol{distance}\) changes. That is what the \(r\) variable is: distance.

Answer 10

decreases by factor of 16

Answer 11

im terrible at physics im confused even how to set this up

Answer 12

it should be .015newton

Answer 13

that isnt a choice^

Answer 14

So you have\[F=k\frac{q_1\ q_2}{r^2}
\]Now increase the distance, \(r\), by a factor of four. Simply, multiply it by 4. That is, put "\(4r\)" in the place of "\(r\)."
That is\[F_{\text{with distance increased by factor of four}}=k\frac{q_1\ q_2}{\left(4r\right)^2}=k\frac{q_1\ q_2}{4^2\ r^2}=k\frac{q_1\ q_2}{16\ r^2}\]

Answer 15

So you have your \[F_{\text{original}}=k\frac{q_1\ q_2}{r^2}
\]and\[F_{\text{with distance increased by factor of four}}=k\frac{q_1\ q_2}{16\ r^2}\]

Answer 16

Actually, let me add to that...\[F_{\text{original}}=k\frac{q_1\ q_2}{r^2}
=.24\ [N]\](The "\([N]\)" is just the units of the force: newtons)

Answer 17

So, the force changed, and that's important. So is there anything you don't understand there?

Answer 18

where do i type this on i dont have a calculator

Answer 19

Well, does the computer you're at have a calculator?

Answer 20

im in a credit retrieval class once im done with this i dont need this anymore I just need this answer and im done I willl try it but honestly i failed like every test in physics to late to learn it all now

Answer 21

And mr.singh was right. As you see, multiplying the distance by four decreases the force by a factor of 16... That's what you'll actually calculate.\[F_{\text{original}}\div 16=F_{\text{with distance increased by factor of four}}\]

Answer 22

Good luck to you, though! I don't give answers, I just help people get to them.

Answer 23

Just what I do :P

Answer 24

But my answer isn't listed... I got\[.24\ [N]\div 16=.015\ [N]\]Are all of the numbers in the problem correct?

Answer 25

im not sure ple.plato.com is sometimes off which should be illegal because it forschool one reason its pissing me off I actually already got that answer but I didnt see it in the choices

Answer 26

Or may I have done something wrong? Would you be able to double-check the problem, @(JPO)666?

Answer 27

@(JPO)666 ?*

Answer 28

It doesn't seem to be linking...

Answer 29

you did it correctly...i got the same answer too..@theEric

Answer 30

ive had questions on there that were supposed to have pics with the question but it ended up just having red x's where the pics were supposed to be this might be wrong or insuffisent data

Answer 31

because they dont seem to give two craps about fixing those problems on plato.com

Answer 32

should i put .012

Answer 33

.12

Answer 34

Huh... That's unfortunate. But those are the words? You might want to talk to your professor to see if you did something wrong or if the site is bad... if possible.

.03 is the closest to .015N....

Answer 35

the sight is bad I already know this I jsut dont know whch ones are faulty

Answer 36

I have to go, I'll be back on in a little bit. Good luck!

Answer 37

ok thanks for your help