What is the Value of F (in F = k1 x q1 x q2/d^2) if d is halved?

richiesmasher

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Hey I have a simple question here, but sometimes its these simple ones that get me.

In a given relationship, the question states: What is the value of (F) if (d) is halved.
I played around with it and I found that F doubles as (d) is halved. I used 2's and 1's in the relationship.
But the answer in my book is this '' F/√2 ''
I have no Idea how they got that.

Here is the relationship: F = k1 x q1 x q2/d2

 

Dr.Peterson

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Actually, neither you nor the book is right. (Are you sure you looked at the answer to the right question?)

How did you get your answer? Can you show the details?
 

mmm4444bot

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… What is the value of (F) if (d) is halved.

F = k1 · q1 · q2 / d2
I don't agree with the book's answer. Let's test it, using the following values.

k1 = 1

q1 = 1

q2 = 1

d = 2

F = (1)(1)(1)/2^2 = 1/4

Now, divide d in half, and see what happens.

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

In this test, halving d resulted in F becoming four times larger.

That's 4F, not F/√2.

Please check to make sure that you've posted everything correctly.

Instead of using trial and error, by substituting values for unknowns, how about we work symbolically, instead?

Starting again with F = k1 · q1 · q2 / d2, replace symbol d with d/2.

F = k1 · q1 · q2 / (d/2)2

Now, simplify and see what you get. 8-)
 

richiesmasher

Junior Member
Joined
Dec 15, 2017
Messages
111
Actually, neither you nor the book is right. (Are you sure you looked at the answer to the right question?)

How did you get your answer? Can you show the details?
I don't agree with the book's answer. Let's test it, using the following values.

k1 = 1

q1 = 1

q2 = 1

d = 2

F = (1)(1)(1)/2^2 = 1/4

Now, divide d in half, and see what happens.

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

In this test, halving d resulted in F becoming four times larger.

That's 4F, not F/√2.

Please check to make sure that you've posted everything correctly.

Instead of using trial and error, by substituting values for unknowns, how about we work symbolically, instead?

Starting again with F = k1 · q1 · q2 / d2, replace symbol d with d/2.

F = k1 · q1 · q2 / (d/2)2

Now, simplify and see what you get. 8-)
In reply to both of you, I'm 100 percent sure my book says F/√2 as the answer.

I got my answer by doing this, I used 2 for the value of everything, so in essence I had ''2 x 2 x2 /2^2 = 8/4, which would give 2. Now when I reach 8/4, I'll now halve (d) and get 2, so I would have 8/2 which would give 4, so I thought my answer doubled.

Now to do it with symbols like you say I would get F = k1 · q1 · q2 / d^2/4 after I simplify.

My book is known for print errors, It's actually a physics textbook as I'm studying that as well, and this formula was for the Force between two charges, but as you can see some maths came up here, so that's how I based my assumptions, kinda confused still but I'm trying.
 

Dr.Peterson

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Joined
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Messages
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I got my answer by doing this, I used 2 for the value of everything, so in essence I had ''2 x 2 x2 /2^2 = 8/4, which would give 2. Now when I reach 8/4, I'll now halve (d) and get 2, so I would have 8/2 which would give 4, so I thought my answer doubled.
But what you halved (the 4) was d^2, not d!

Halving d would mean replacing 2 with 1 in the original: 2 x 2 x2 /1^2 = 8/1 = 8, which is 4 times what it was.

Now to do it with symbols like you say I would get F = k1 · q1 · q2 / d^2/4 after I simplify.
Keep going! This is a complex fraction; to simplify it, you can multiply the numerator and denominator by 4, to get F = 4 k1 · q1 · q2 / d^2. But this is 4F. In halving d, you have multiplied F by 4.
 

richiesmasher

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Dec 15, 2017
Messages
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But what you halved (the 4) was d^2, not d!

Halving d would mean replacing 2 with 1 in the original: 2 x 2 x2 /1^2 = 8/1 = 8, which is 4 times what it was.



Keep going! This is a complex fraction; to simplify it, you can multiply the numerator and denominator by 4, to get F = 4 k1 · q1 · q2 / d^2. But this is 4F. In halving d, you have multiplied F by 4.
Ok, I'm going out for a bit now, when I return I definitely will be tackling this.
 

richiesmasher

Junior Member
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Dec 15, 2017
Messages
111
You can get rid of those 3 funny useless variables by
assigning 1 to each (as Mark did) to get:
F = 1/d^2 [1]
Then:
F = 1/(d/2)^2

F = 1/(d^2 / 4)

F = 4 / d^2 [2]

[2] = [1] * 4
But what you halved (the 4) was d^2, not d!

Halving d would mean replacing 2 with 1 in the original: 2 x 2 x2 /1^2 = 8/1 = 8, which is 4 times what it was.



Keep going! This is a complex fraction; to simplify it, you can multiply the numerator and denominator by 4, to get F = 4 k1 · q1 · q2 / d^2. But this is 4F. In halving d, you have multiplied F by 4.
Ah yes this clears it up fully, thank you guys, now the mystery has been solved and I can be at peace. I wasn't aware about the complex fractions, as I noted sometimes it's these fundamental things that get me, but it truly showed me that it's better to do things symbolically first, to see what the final expression would be, then substitute values, I didn't even think of that before but now it makes good sense to me.
 
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