Please help me with this physics questions asap

Star34

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#1
Two point charges, initially 2.0 cm apart, experience a 1.0-Nforce. If they are moved to a new
separation of 8.0 cm, what is the electric force between them?
 

Star34

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#2
Please dont make fun of me for posting here, instead if youre good in physics try to help me, thanks.
 

Star34

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#4
@Kaspar i tried but im getting the wrong amswer, can u please do it if you have time if not it's fine, and im really thankful for your help
 

Kaspar

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#6
Ok so before I help you, how much do you know? What do you know about force of attraction/repulsion, inversely proportional and directly proportional? To understand this physic's exercise, you'll need to know about those terms...
 

Star34

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#7
this is how i tried to solve it.
P.s it doest allow me to post pics saying waiting for mods
 

Kaspar

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#9
Okey mate. I'll explain it to you.

"Coulomb's law states that:
The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them."


In this case, the force is inversely proportional to the distance squared. The charges are not changed. The distance is altered from 2.0 cm apart to 8.0 cm apart (4 times as far, which shows that the force is a ratio of 1/16). The initial force was 1,0N. Multiply it with 1/16 and you get the answer: 1,0N*(1/16) = 0,0625N.

That's one way to solve it just by using your knowledge and coming to that conclusion.
Here's another way.
Since the force is inversely proportional to the distance squared, we can conclude that: F1/F2 = (r2^2)/(r1^2)
F1 = 1,0N
r1 = 2,0cm = 0,02m
r2 = 8,0cm = 0,08m
Add the values to the equation:
1/F2 = (0,08^2)/(0,02^2)
1 = F2*(0,08^2)/(0,02^2)
F2 = 1/((0,08^2)/(0,02^2))
F2 = 0,0625N

Good luck with your studies!
 

Star34

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#10
@Kaspar thanks bro, for your explanation youre better than my physics teacher your way is organized and easily understandable.
 
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