Consider three charges, $q_{1} = q_{2} = q_{3} = q$ $q_1=q_2=q_3=q$ , at the vertices of an equilateral triangle of side $l$ $l$ . What is the force on a charge $Q$ $Q$ (with the same sign as $q$ $q$ ) placed at the centroid of the triangle? Solve using parallelogram law of vectors.

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Answer 1 · 2018-03-16T18:07:08

The force is zero.

$sf(F=k.(q_1q_2)/(r^2))$

$sf(F)$ is the force.

$sf(k )$ is the Coulombic Constant.

$sf(q_1)$ and $sf(q_2)$ are the charges.

$sf(r)$ is the separation.

An equilateral triangle looks like this:

![keywordsuggest.org]( useruploads.socratic.org )

$sf(h_a=h_b=h_c)$

Since all the charges are equal and the distance between them is the same the forces look like this:

MFDocs

From the symmetry of the situation you can see that the net force must be zero.

If you wish to show this you can set Q to be at the origin.

You can resolve $sf(F_1)$ and $sf(F_3)$ to give $sf(F_(res))$ .

Because we have an equilateral triangle then $sf(theta=60^@)$ .

$:.$ $sf(F_(res)=F_1costheta+F_3costheta)$

Since $sf(F_1=F_3=F)$ then:

$sf(F_(res)=2Fcostheta=2Fxx0.5=F)$

This is acting upwards. It is exactly balanced by $sf(F_2)$ acting downwards. Since $sf(F_2=F)$ then the net force on Q will be $sf(F-F=0)$

Consider three charges, q_1=q_2=q_3=qq1=q2=q3=qq_1=q_2=q_3=q, at the vertices of an equilateral triangle of side lll. What is the force on a charge QQQ (with the same sign as qqq) placed at the centroid of the triangle? Solve using parallelogram law of vectors.