Answer:
Therefore the magnitude of the net electric filed at the center of the triangle is [tex]=\frac{6K}{A^2}[/tex] N/C
Explanation:
Given,A = side of the triangle. q = 1 C
The center of a triangle is the centroid of the triangle.
The distance between centroid to any vertices of a equilateral triangle is
[tex]=\frac{2}{3}[/tex] of the height of the equilateral triangle
[tex]=\frac{2}{3}\times \frac{\sqrt{3} }{2} \times A[/tex]
[tex]=\frac{1}{\sqrt{3} } A[/tex]
Electric field= [tex]\frac{Kq}{d^2}[/tex] K=8.99×10⁹ Nm²/C², q=charge and d = distance
Therefore the magnitude of the net electric filed at the center of the triangle is
=2[tex]\frac{Kq}{d^2}[/tex] [both charge are at same distance from the centroid]
=[tex]\frac{2k}{(\frac{1}{\sqrt{3} }A)^2 }[/tex]
[tex]=\frac{6K}{A^2}[/tex] N /C [K=8.99×10⁹ Nm²/C²]
