Show that this equation is inverse square

  • Thread starter Thread starter brycenrg
  • Start date Start date
  • Tags Tags
    Inverse Square
AI Thread Summary
The equation F = k(Q2*Q1)/(r^2) represents an inverse square law, where the force F is inversely proportional to the square of the distance r between two charges Q1 and Q2. The constant k remains unchanged while varying r allows for observing the relationship. Users are advised to keep Q1 and Q2 constant and only change r to see how it affects the force. This approach clarifies that as r increases, the force decreases according to the inverse square relationship. Understanding this principle is crucial for grasping the underlying physics of the equation.
brycenrg
Messages
95
Reaction score
2

Homework Statement


F = k(Q2*Q1)/(r^2)

Homework Equations

The Attempt at a Solution


I asked my teacher and he said that this is an inverse square law. Didn't say anything else. He also mentioned k is constant.

I assume i can plug in random values and see if there is a pattern... k=1 for all
Set 1, q1 = 2, q2 = 4, r = 5
F1 = 8/25
Set 2, q1 = 1 q2 = 5, r = 4
F2 = 5/16
I plugged in random values but i don't really get it.
What is inverse of what? I see that the r is squared and its below the kq1q2. I'm stuck.

Any advice is appreciated.
 
Physics news on Phys.org
This function is dependent on the inverse squared of r, which is usually the distance between the locations of Q1 and Q2.
 
brycenrg said:

Homework Statement


F = k(Q2*Q1)/(r^2)

Homework Equations

The Attempt at a Solution


I asked my teacher and he said that this is an inverse square law. Didn't say anything else. He also mentioned k is constant.

I assume i can plug in random values and see if there is a pattern... k=1 for all
Set 1, q1 = 2, q2 = 4, r = 5
F1 = 8/25
Set 2, q1 = 1 q2 = 5, r = 4
F2 = 5/16
I plugged in random values but i don't really get it.
What is inverse of what? I see that the r is squared and its below the kq1q2. I'm stuck.

Any advice is appreciated.
What RUber said . ...

So, pick values for Q1 and Q2 and stay with those. Then plug-in various values for r.
 
  • Like
Likes brycenrg
SammyS said:
What RUber said . ...

So, pick values for Q1 and Q2 and stay with those. Then plug-in various values for r.
Thank you so, r is the only value that is changing?
 
Ray Vickson said:
Yes. That is exactly what "inverse square" means. Google is your friend; see
https://en.wikipedia.org/wiki/Inverse-square_law
Thank you, Yeah i googled it but thought the Q's changed as well. So it didn't make sense. Sometimes i need someone to explain it to me like I am 5.
 
  • Like
Likes RUber
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top