The magnitude of the electric force on a 2, m | vedclass

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Question

(D) The distance from the center $O$ to each vertex of the equilateral triangles is $r = \frac{a}{\sqrt{3}} = \frac{10\, cm}{\sqrt{3}} = \frac{0.1}{\s

Vedclass · Accepted Answer

$43.2$

Vedclass · Answer

(D) The distance from the center $O$ to each vertex of the equilateral triangles is $r = \frac{a}{\sqrt{3}} = \frac{10\, cm}{\sqrt{3}} = \frac{0.1}{\sqrt{3}}\, m$.Let $q = 2\, \mu C = 2 	imes 10^{-6}\, C$ be the charge at the center $O$ and $q_0 = 2\, \mu C$ be the magnitude of the charges at the vertices.The force due to each charge at the center is $F = \frac{k q q_0}{r^2} = \frac{9 	imes 10^9 	imes (2 	imes 10^{-6})^2}{(0.1/\sqrt{3})^2} = \frac{9 	imes 10^9 	imes 4 	imes 10^{-12}}{0.01/3} = 36 	imes 10^{-3} 	imes 300 = 10.8\, N$.Looking at the symmetry,the forces from $A, B, C$ (positive) and $D, E, F$ (negative) act on the central charge $q$. Charges $A, B, C$ attract the negative charge or repel the positive charge. Since the central charge is positive $(2\, \mu C)$,it is repelled by $A, B, C$ and attracted by $D, E, F$.The vectors of these forces point towards the vertices $D, E, F$ and away from $A, B, C$. Summing these vectors,we find the resultant force $F_{net} = 4F = 4 	imes 10.8 = 43.2\, N$.

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