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

[Mike94]

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 !

 

[Gordianus]

Do you know something about Coulomb's law?

 

[Mike94]

Do you know something about Coulomb's law?

What do you mean by that ?

 

[Gordianus]

If you have charged bodies you'll notice electric forces. Mr. Coulomb taught us about them.

 

[Mike94]

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.

 

[haruspex]

What I found till now is that F2,1 is 0.081.

Please explain how you get that.

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?

 

[Mike94]

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

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 !

 

[haruspex]

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.

 

[Mike94]

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 !

 

[haruspex]

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.

 

[Mike94]

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.

 

[haruspex]

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?

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.

 

[Mike94]

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.

 

[haruspex]

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

 

[Mike94]

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.

 

[haruspex]

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?

 

[Mike94]

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

 

[haruspex]

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

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.

 

[Mike94]

That's how I get it the numbers.
Thank you.

 

[haruspex]

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 $(d+2)^2=9d^2$

 

[Mike94]

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 !