Electric charges of 1,mu C, -1,mu C, and 2,mu | vedclass

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(B) The side length $r = 10\,cm = 0.1\,m$. The charges are $q_A = 1 	imes 10^{-6}\,C$, $q_B = -1 	imes 10^{-6}\,C$, and $q_C = 2 	imes 10^{-6}\,C$.

Vedclass · Accepted Answer

$1.8$

Vedclass · Answer

(B) The side length $r = 10\,cm = 0.1\,m$. The charges are $q_A = 1 	imes 10^{-6}\,C$, $q_B = -1 	imes 10^{-6}\,C$, and $q_C = 2 	imes 10^{-6}\,C$.The force $F_A$ on $C$ due to charge at $A$ is repulsive:$F_A = \frac{k |q_A q_C|}{r^2} = \frac{9 	imes 10^9 	imes 10^{-6} 	imes 2 	imes 10^{-6}}{(0.1)^2} = 1.8\,N$.The force $F_B$ on $C$ due to charge at $B$ is attractive:$F_B = \frac{k |q_B q_C|}{r^2} = \frac{9 	imes 10^9 	imes 10^{-6} 	imes 2 	imes 10^{-6}}{(0.1)^2} = 1.8\,N$.The angle between the force vectors $F_A$ and $F_B$ is $120^\circ$ because the internal angle of the equilateral triangle is $60^\circ$ and the vectors are oriented such that the angle between them is $180^\circ - 60^\circ = 120^\circ$.The resultant force $F_{net}$ is given by:$F_{net} = \sqrt{F_A^2 + F_B^2 + 2 F_A F_B \cos(120^\circ)}$$F_{net} = \sqrt{(1.8)^2 + (1.8)^2 + 2(1.8)(1.8)(-0.5)}$$F_{net} = \sqrt{1.8^2 + 1.8^2 - 1.8^2} = 1.8\,N$.

Electric charges of $1\,\mu C$, $-1\,\mu C$, and $2\,\mu C$ are placed in air at the corners $A$, $B$, and $C$ respectively of an equilateral triangle $ABC$ having a side length of $10\,cm$. The resultant force on the charge at $C$ is......$N$

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