# In triangle charge q at three corner than electric field at the center of the triangle is zero how explain

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In a triangle there are many centers, Incenter, circum center, center of gravity - centroid, and so on.... The centroid is not at the same distance from the three vertices.. so is incenter.. incenter is equidistant from the sides...

I suppose the given triangle is an equilateral triangle ABC with mid points of the sides AC, AB and BC being D, E and F respectively. The medians BD, CE and AF intersect at G.

Since that is an equilateral triangle, the center is the centroid G.

The three medians have equal length. so GA = GB = GC. So the magnitude of the electric fields at the center due to charges at A, B and C are equal..

Now we prove by vector method that vectors

the following are all vectors..

AG = 2/3 (AB + BC/2)

BG = 2/3 (BC+CA/2)

CG = 2/3(CB+BA/2)

adding : AG + BG + CG = 2/3 (AB+BC+CB) + 1/3(BC+CA+BA)

= 0

since resultant field at G = K * Q / |AG|^2 * [ AG + BG + CG ]

that is 0.

I think that this is only good for an equilateral triangle.

I suppose the given triangle is an equilateral triangle ABC with mid points of the sides AC, AB and BC being D, E and F respectively. The medians BD, CE and AF intersect at G.

Since that is an equilateral triangle, the center is the centroid G.

The three medians have equal length. so GA = GB = GC. So the magnitude of the electric fields at the center due to charges at A, B and C are equal..

Now we prove by vector method that vectors

the following are all vectors..

AG = 2/3 (AB + BC/2)

BG = 2/3 (BC+CA/2)

CG = 2/3(CB+BA/2)

adding : AG + BG + CG = 2/3 (AB+BC+CB) + 1/3(BC+CA+BA)

= 0

since resultant field at G = K * Q / |AG|^2 * [ AG + BG + CG ]

that is 0.

I think that this is only good for an equilateral triangle.