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Let k = 1/4π€°
The electric field intensity (E) at O due to A is |E1 |= k (q/r^2) [directed towards midpoint of BC]——-(1)
And electric field intensity (E) at O due to B is |E2| = k (q/r^2) [directed towards midpoint of AC] ———(2)
The electric field intensity (E) at O due to C is |E3| = k (q/r^2) [directed towards midpoint of AB] ——-(3)
From (1),(2),(3) we get the net electric intensity at O due to A,B,C combined is given by ,
E = E1 + E2 + E3
Note that electric field intensity E1,E2,E3 are vector quantities and they form a triad with angle between them = 120°, and since they are equal in magnitude they cancel out each other
Thus ,net electric intensity
E = 0
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