Three equal charges are situated at the vertex of an equilateral

In summary, Delphi51 found that q'\approx{1.75Q} is the correct charge for the center of the equilateral triangle.
  • #1
Telemachus
835
30
Hi there. I did this problem, and I wanted to know if my solution is ok.

The problem says: Three equal charges are situated at the vertex of an equilateral triangle, and we want to cancel out the forces exerted by each other. What charge q' of opposite sign should be placed at the center of the triangle?

I've made a picture:
attachment.php?attachmentid=38630&stc=1&d=1315404825.png


The side of the triangle: [tex]l=2r\cos 30º=\sqrt[ ]{3}r[/tex]
And the height: [tex]h=r+r\sin 30º=\displaystyle\frac{3}{2}r[/tex]

Then I set the force summatory for Q2.

[tex]\vec{F_2}=\displaystyle\frac{Q_2}{4\pi \epsilon_0}\left ( \frac{4Q_1}{9r^2}\hat{j}+\frac{4Q_1}{3r^2}\hat{i}+\frac{Q_3}{r^2}\hat{i}-\frac{4q'}{r^2}\hat{j}-\frac{4q'}{3r^2}\hat{i} \right)[/tex]

Under the conditions:
[tex]\vec{F_2}=0,Q_1=Q_3=Q[/tex]

I get:
[tex]q'\hat{i}=\displaystyle\frac{7}{4}Q\hat{i},q'\hat{j}=\displaystyle\frac{Q}{9}j[/tex]

Then [tex]||q'||\approx{}1.75Q[/tex]

And then: [tex]q'\approx{1.75Q}[/tex]

Is this right?
 

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  • #2


Curious, I get a slightly different answer and haven't the patience to check through your 2D calc. Consider this ...
The force on Q2 due to Q1 is F = kQ²/(3r²).
Q3 on Q2 is the same and components perpendicular to q'Q2 cancel, so double and use cos(30) to get the components parallel to q'Q2 (left side), which must be equal to the q'Q2 force (right side):
2*kQ²/(3r²)*cos(30) = kQq'/r²
just cancelling ...
2/3*Q*cos(30) = q'
0.577*Q = q'
 
  • #3


Thank you very much Delphi. Is something wrong about my reasoning? without getting into details on my decomposition in Cartesian coordinates. What I did was, from the equation for the summatory of forces:

[tex]\displaystyle\frac{4Q_1}{9r^2}\hat{j}+\frac{4Q_1}{3r^2}\hat{i}+\frac{Q_3}{r^2}\hat{i}-\frac{4q'}{r^2}\hat{j}-\frac{4q'}{3r^2}\hat{i}=0[/tex]

Then, separating into the different components, and computing the modulus I get into my result.

I want to know if my conception was right, it doesn't matter if I made a mistake when making the decomposition over i and j, but I do care if I reasoned this wrong.
 
  • #4


I think you might have a problem somewhere in here:
Then I set the force summatory for Q2.

[tex]\vec{F_2}=\displaystyle\frac{Q_2}{4\pi \epsilon_0}\left ( \frac{4Q_1}{9r^2}\hat{j}+\frac{4Q_1}{3r^2}\hat{i}+\frac{Q_3}{r^2}\hat{i}-\frac{4q'}{r^2}\hat{j}-\frac{4q'}{3r^2}\hat{i} \right)[/tex]


At Q2, you can argue that the forces in the x direction must equal 0, as well as the forces in the y direction. So, taking just the x direction:
[tex]\vec{F_2x}=\displaystyle\frac{Q_2}{4\pi \epsilon_0}\left ( \frac{-cos(60)Q_1}{3r^2}\hat{i}-\frac{Q_3}{3r^2}\hat{i}+\frac{cos(30)q'}{r^2}\hat{i} \right)[/tex]

[tex]0=\displaystyle \frac{-cos(60)Q}{3}-\frac{Q}{3}+\frac{cos(30)q'}{1} [/tex]

which results in the answer Delphi51 got (which is 1/3rd of what you got).
 
  • #5


Thank you very much sir :), I'll review this later with more time (I have to go to school right now) but I think you got it.
 

1. What is the force between the three equal charges at the vertex of an equilateral triangle?

The force between the three charges can be calculated using Coulomb's law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

2. How do the charges affect the electric field at the center of the equilateral triangle?

The electric field at the center of the equilateral triangle is affected by the charges in a way that the resultant electric field is zero. This is because the electric field vectors from each charge will cancel out due to the symmetry of the equilateral triangle.

3. What is the potential energy of the system of three equal charges at the vertex of an equilateral triangle?

The potential energy of the system can be calculated using the formula U = (k * q^2)/a, where k is the Coulomb's constant, q is the magnitude of the charges, and a is the distance between the charges.

4. Can the charges at the vertex of an equilateral triangle be of any magnitude?

Yes, the charges can be of any magnitude as long as they are equal. The force and potential energy between the charges will vary depending on the magnitude of the charges, but the overall electric field at the center of the triangle will still be zero.

5. How would the system of three equal charges at the vertex of an equilateral triangle change if one of the charges is removed?

If one of the charges is removed, the remaining two charges will still form an equilateral triangle, but the force and potential energy between them will change. The electric field at the center of the triangle will also be affected, but it will still be zero due to the symmetry of the remaining charges.

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