Three positive point charges are located on a line (Coulomb's Law)

In summary: I meant F2,1. F2 is the Coulomb's constant, not the resultant force.F2,1 is Coulomb's constant, not the resultant force.
  • #1
Mike94
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0
Thread moved from the technical forums, so no Homework Template is shown.
Homework Statement
Q2 = 2 Microcoulombs Q3 = 6 Microcoulombs

The distance between Q2 and Q3 is 2 meters


Q1(+)-------------Q2(+)-------------Q3(+)


If the charge of Q2 is subjected to a resultant force of 54⋅E-3 N to the right, and the magnitude of the resultant force on Q1 is twice as large as the resultant force on Q3, determine the value of Q1 and the distance between Q1 & Q2
Relevant Equations
What I found till now is that F2,1 is 0.081.

Equations that I have, but I still can't resolve q1 and d

F2,1+F2,3=F2

F1,2+F1,3=−2*(F3,1+F3,2)
I need some help resolving the follow problem. I really don't know where to put the "twice as large as the resultant force on Q3" in order to build an equation.

Thank you !
 
Last edited:
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  • #2
Do you know something about Coulomb's law?
 
  • #3
Gordianus said:
Do you know something about Coulomb's law?

What do you mean by that ?
 
  • #4
If you have charged bodies you'll notice electric forces. Mr. Coulomb taught us about them.
 
  • #5
So, since F2,1 is 0.081
E2,1 = k|q2| / r^2
F2,1 = q1 * E2,1

Am I on the right track ?

Thanks.
 
  • #6
Mike94 said:
What I found till now is that F2,1 is 0.081.
Please explain how you get that.
Mike94 said:
Equations that I have, but I still can't resolve q1 and d
F2,1+F2,3=F2
F1,2+F1,3=−2*(F3,1+F3,2)
where to put the "twice as large as the resultant force on Q3" in order to build an equation.
Isn't that what you have expressed in the second equation above?
F2 is known, so those two equations have three unknowns.
You have not used the distance between q2 and q3. What can you get from that?
 
  • #7
In order to get F2,1 = 0.081

I did:

F2,1 + F2,3 = 0.054
F2,1 + -(F2,3 = 0.027 by using the Coulomb's law) = 0.054
So F2,1 = 0.081
haruspex said:
You have not used the distance between q2 and q3. What can you get from that?

I guess the distance ?

Can I do F1 = F1,2 + F1,3
Set F1 to 0 and solve F1,2 = F1,3 in order to get the distance ?

Thanks !
 
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  • #8
Mike94 said:
In order to get F2,1 = 0.081

I did:

F2,1 + F2,3 = 0.054
F2,1 + -(F2,3 = 0.027 by using the Coulomb's law) = 0.054
So F2,1 = 0.081
Ah, so you used the distance between 2 and 3 to find F2,3. Ok.
You have
F1,2+F1,3=−2*(F3,1+F3,2)
Where F1,2=-F2,1 etc., and F2,1 and F2,3 are known. So that's enough to find all the forces.
Now you need equations relating q1 and the distance from q1 to q2 to these forces.
 
  • #9
haruspex said:
Now you need equations relating q1 and the distance from q1 to q2 to these forces.

Not sure how about those equations, the only thing I see is:

F1,2 = KQ1Q2 / -d^2
F1,3 = KQ1Q3 / (-d+2)^2

I really don't see another way to find equations related to the distance. If you can give me a hint to the right direction.

Thanks again !
 
  • #10
Mike94 said:
Not sure how about those equations, the only thing I see is:

F1,2 = KQ1Q2 / -d^2
F1,3 = KQ1Q3 / (-d+2)^2

I really don't see another way to find equations related to the distance. If you can give me a hint to the right direction.

Thanks again !
I don't understand the minus signs on the d's.
Other than that, you have two equations with only two unknowns, d and q1.
 
  • #11
haruspex said:
I don't understand the minus signs on the d's.

Minus signs is because Q2 & Q3 are pushing the Q1 to the left.

KQ1Q2 / d^2 = KQ1Q3 / (d+2)^2
Which d = 2.73 and q1 = 0.000011 ?

Thanks.
 
  • #12
Mike94 said:
Minus signs is because Q2 & Q3 are pushing the Q1 to the left.
Ok, so F1,2 = KQ1Q2 / -d^2 was just a n unusual way of writing F1,2 = -KQ1Q2 / d^2, but F1,3 = KQ1Q3 / (-d+2)^2?
You meant F1,3 = -KQ1Q3 / (d+2)^2, right?
Mike94 said:
KQ1Q2 / d^2 = KQ1Q3 / (d+2)^2
How do you get that? You seem to be assuming F1,3=F1,2. You already found all the forces (see post #8), just plug in F1,3 and F1,2 from there.
 
  • #13
haruspex said:
You meant F1,3 = -KQ1Q3 / (d+2)^2, right?
That's right.

So, I can do:

F1,2 = kQ1Q2 / d^2
F1,3 = -kQ1Q3 / (d+2)^2
F3,1 = kQ3Q1 / (d+2)^2
F3,2 = kQ3Q2 / 2^2

Doing a solve on F1,2 + F1,3 = - 2 (F3,1 + F32) to find q1 & d
I'm surely missing something since the calculator cannot resolve.

Thanks.
 
  • #14
Mike94 said:
That's right.

So, I can do:

F1,2 = kQ1Q2 / d^2
F1,3 = -kQ1Q3 / (d+2)^2
F3,1 = kQ3Q1 / (d+2)^2
F3,2 = kQ3Q2 / 2^2

Doing a solve on F1,2 + F1,3 = - 2 (F3,1 + F32) to find q1 & d
I'm surely missing something since the calculator cannot resolve.

Thanks.
In your post #1 you already had enough to find all the forces. Just use the values you get from those for F1,2 and F1,3 in the equations
F1,2 = -kQ1Q2 / d^2
F1,3 = -kQ1Q3 / (d+2)^2
 
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  • #15
Thank you a lot!

I think I got it.

Q1 = 0.000238
D = 8.89898

By doing

-k*q2*0.000238 / 8.89898^2 + k*q2*q3 / 2^2 = 0.081 , exactly what F2 is.
 
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  • #16
Mike94 said:
Thank you a lot!

I think I got it.

Q1 = 0.000238
D = 8.89898

By doing

-k*q2*0.000238 / 8.89898^2 + k*q2*q3 / 2^2 = 0.081 , exactly what F2 is.
I don't understand. I thought F2 was given. The 0.081 figure is F2,1, no?
What units are you expressing q1 in? Shouldn't the answer be similar to the others, a few μC?
 
  • #17
haruspex said:
I don't understand. I thought F2 was given. The 0.081 figure is F2,1, no?
What units are you expressing q1 in? Shouldn't the answer be similar to the others, a few μC?
Sorry, I was happy for nothing. Exhausted trying all day long to understand this.

What I did was:

kQ2Q1/d^2 + -kQ2Q3 / 2^2 = 0.054
F1,2+F1,3=-2*(F3,1+F3,2)

Q1= 0.000005 & d =1 or Q1=0.000001 and d = -0.5

Which aren't the right numbers because F2,1 + F2,3 != 0.054
 
  • #18
Mike94 said:
kQ2Q1/d^2 + -kQ2Q3 / 2^2 = 0.054
F1,2+F1,3=-2*(F3,1+F3,2)
Yes, that's well established, but how do you get from there to
Mike94 said:
Q1= 0.000005 & d =1 or Q1=0.000001 and d = -0.5
?
I cannot tell where you are going wrong if you do not show all your steps.
 
  • #19
Screenshot_25.png


That's how I get it the numbers.
Thank you.
 
  • #20
Mike94 said:
View attachment 270426

That's how I get it the numbers.
Thank you.
Not sure what tool you are using, but it seems to be giving you a rounding error in the value of q1. It should be 4.5μC.
You don't really need to use such a tool.
You have ##\frac{2q}{d^2}-\frac{2\times 6}{2^2}=\frac{54000}k## where k=9000, and
##\frac{2q}{d^2}+\frac{6q}{(d+2)^2}=2(\frac{6q}{(d+2)^2}+\frac{2\times 6}{2^2})##
This might look messy but it quickly comes down to ##2q=9d^2## and thence ##(d+2)^2=9d^2##
 
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  • #21
haruspex said:
Not sure what tool you are using, but it seems to be giving you a rounding error in the value of q1. It should be 4.5μC.

I'm using Ti-Nspire . I will use another calculator.

Thanks a lot & have a wonderful day !
 

Related to Three positive point charges are located on a line (Coulomb's Law)

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law in physics that describes the electrostatic force between two charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. How is Coulomb's Law applied to three positive point charges on a line?

In the case of three positive point charges on a line, Coulomb's Law can be used to calculate the force between each pair of charges. The total force on each charge is then the vector sum of the individual forces.

3. What is the formula for Coulomb's Law?

The formula for Coulomb's Law is F = k * (q1 * q2) / r^2, where F is the force between two charges, k is the Coulomb's constant (8.99 x 10^9 N*m^2/C^2), q1 and q2 are the charges of the two particles, and r is the distance between them.

4. How does the distance between the charges affect the force according to Coulomb's Law?

According to Coulomb's Law, the force between two charges is inversely proportional to the square of the distance between them. This means that as the distance between the charges increases, the force decreases. Similarly, as the distance decreases, the force increases.

5. Can Coulomb's Law be used to calculate the force between any two charged particles?

Yes, Coulomb's Law can be used to calculate the force between any two charged particles, as long as the particles are point charges and the distance between them is large compared to their size. It is a fundamental law in electrostatics and is applicable to all charged particles, regardless of their size or composition.

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