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Heaps 3, Q83

Heaps 3, Q83

Postby skogias6733 » Fri Apr 08, 2022 6:05 pm

Hi, i am having a little trouble understanding the maths in this question

Q83. if two particles retain their charge (q), by what distance must they now be separated in order for the force between to increase by 2?

Worked Solution : Discuss this question
Using Coulomb's equation : F = kq1q2/r2
At original distance, f = kq1q2/r2

r = √(kq1q2/f)
At new distance x, 2f = kq1q2/x2

x = √(kq1q2/2f)

x = (1/√2) x √(kq1q2/f) = (1/√2) x r = r/√2


i understand up to x = 1/√2
im assuming k was given the value of 1 and q1 and q2 are left out because theyre constant, so that

x = √(kq1q2/2f)

But then why is the 1/√2 then multipled by √(kq1q2/f) ??
and then the final value is 1/√2 or r/√2 but doenst this mean the answer would be 0.7?

1/√2 = 0.7 and √1/2 = 0.7 so why is the answer for distance (x or r) then just √2 on its own?

Thanks in advance!
skogias6733
 
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Re: Heaps 3, Q83

Postby goldstanda3269 » Fri Apr 08, 2022 9:56 pm

skogias6733 wrote: i understand up to x = 1/√2
im assuming k was given the value of 1 and q1 and q2 are left out because theyre constant, so that


Not exactly. We'll try it a different way.

Using Coulomb's equation : F = kq1q2/r2

In the equation, everything will stay the same except F and r (we do not know what k, q1 and q2 are, but we know that they are constant). Both sides of the equation must change in a way that assures that the equation remains balanced. We are told that the force doubles thus we can rewrite the equation this way:

2F = 2(kq1q2/r2)

So far, so good! BUT, how is r squared in the denominator changed by 2 in the numerator? How can we get that number 2 so that it is sitting right next to r AND it is within the square just like r is?

Well, there is an expression that we use in the GAMSAT Maths chapters: a denominator in the denominator is the same as a numerator. In other words:

x = 1 / (1/x)

(you might cross multiply or have some other way to remember)

2(kq1q2/r2) = (kq1q2)/(r2)/2

Now to get the number 2 within the square, just like r, in order to see how r is affected, we need to take the square root of 2:

(kq1q2)/(r2)/2 = (kq1q2)/(r/root 2)2

What did we prove? If you multiply F by 2, it has the same effect as dividing r by root 2. Thus, a fancy way to say it: "The distance between them must be decreased by a factor of √2."


There are many, many examples of equation manipulations in the Masters Series: Please consider looking at the following Physics video... New: Simple Pendulum and Practice Problems.
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Re: Heaps 3, Q83

Postby skogias6733 » Sat Apr 09, 2022 10:33 am

Hi,

Thankyou for that!

Can we also say that since r = square root of k(q1q2)/F
and since kq1q2 dont change they are left out and since force is now 2F,
r is just square root of 2F? ( or is it improper leave them out if theyre constant? )

or that since Force increases by 2 and since force and distance follow an 'inverse squared law' that the distance would decrease by the inverse of squared function = square root ?

And so the 2 in the square root comes from the fact that the RHS of equation had to be multiplied by 2 as well?

Im aware that you have to do things to both sides in manipulating equations, but i didnt realise we had to in this question i thought it just follows the inverse square law, like how if distance is doubled, then force is quartered because of that inverse square relationship so that if force is doubled, then the distance decreases by the inverse of the squared function.

i have completed masters series but i am still trying to improve my maths/physics weakness .
Appreciate the help thank you!!
skogias6733
 
Posts: 35
Joined: Sat Feb 06, 2021 11:17 am

Re: Heaps 3, Q83

Postby goldstanda3269 » Tue Apr 12, 2022 6:39 pm

skogias6733 wrote:Hi,

Thankyou for that!

Can we also say that since r = square root of k(q1q2)/F
and since kq1q2 dont change they are left out and since force is now 2F,
r is just square root of 2F?


Essentially, correct.

Technically, it would be said this way regarding this particular problem: r is proportional to the square root of 2F.


skogias6733 wrote:H

or that since Force increases by 2 and since force and distance follow an 'inverse squared law' that the distance would decrease by the inverse of squared function = square root ?

And so the 2 in the square root comes from the fact that the RHS of equation had to be multiplied by 2 as well?


That is also fair to say.
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