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A point P is located at a height h = l / (2*sqrt(6)) above the centroid of an equilateral triangle with side length l and uniform surface charge density sigma. The electric field at P is sigma / (alpha * epsilon₀). Find the value of alpha.

1. 2*sqrt(6)
2. 3*sqrt(6)
3. 6*sqrt(6)
4. 12*sqrt(6)

Correct answer: 6*sqrt(6)

Solution

For a uniformly charged equilateral triangle with surface charge density sigma and side l, the electric field at a point on the axis (perpendicular to the plane through the centroid) at height h is found by integration. The circumradius of the equilateral triangle is R_c = l/sqrt(3). Given h = l/(2*sqrt(6)), the field works out to E = sigma/(6*sqrt(6)*epsilon₀). So alpha = 6*sqrt(6).

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