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Consider three charges \(q_1,q_2,q_3\) each equal to \(q\) at the vertices of an equilateral triangle of side \(l\) . What is the force on a charge \(Q\) \((\)with the same sign as \(q)\) placed at the centroid of the triangle, as shown in figure.

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Consider three charges \(q_1,q_2,q_3\) each equal to \(q\) at the vertices of an equilateral triangle of side \(l\) . What is the force on a charge \(Q\) \((\)with the same sign as \(q)\) placed at the centroid of the triangle, as shown in figure.
Question: Consider three charges  q_1,q_2,q_3  each equal to  q  at the vertices of an equilateral triangle of side  l . What is the force on a charge  Q (with the same sign as  q) placed at the centroid of the triangle, as shown in figure.

Answer

The equilateral triangle \(\mathit{ABC}\)  given, has equal length of each side \(l\) , and if we draw a perpendicular line \(\mathit{AD}\)  to the side \(\mathit{BC}\) , then
\(\mathit{AD}=\mathit{AC}\cos 30^0=(\sqrt 3/2)l\)
and the distance \(\mathit{AO}\) of the centroid \(O\)  from \(A\)  is
\((2/3\)
And by the law of symmetry, we know that \(\mathit{AO}=\mathit{BO}=\mathit{CO.}\)
Thus,
Force \(F_1\) acting on \(Q\)  due to the charge \(q\)  at \(A=(3/4\pi \varepsilon _0)\times (\mathit{Qq}/l^2)\) along dotted line \(\mathit{AO.}\)
Force \(F_2\) acting on \(Q\)  due to the charge \(q\)  at \(B=(3/4\pi \varepsilon _0)\times (\mathit{Qq}/l^2)\) along dotted line \(\mathit{BO.}\)
Force \(F_3\) acting on \(Q\)  due to the charge \(q\)  at \(C=(3/4\pi \varepsilon _0)\times (\mathit{Qq}/l^2)\) along dotted line \(\mathit{CO.}\)
Resultant of forces \(F_2\) and \(F_3=(3/4\pi \varepsilon _0)\times (\mathit{Qq}/l^2)\) along \(\mathit{OA}\)  by the parallelogram law.
Thus, the total force \(F\) on charge \(Q\) can be given as,
\(F=\frac 3{4\pi \varepsilon _0}\frac{\mathit{Qq}}{l^2}\left(r-r\right)=0\)
Where \(r\) is the unit vector along dotted line \(\mathit{OA.}\)
Also it is revealed by symmetry that the three forces willobviously sum to zero.
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