Hi there. I did this problem, and I wanted to know if my solution is ok.(adsbygoogle = window.adsbygoogle || []).push({});

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:

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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# Homework Help: Three equal charges are situated at the vertex of an equilateral

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