Three charges of each magnitude 100 mu C are | vedclass

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(A) Let the charges be $q_A = +100 \mu C$,$q_B = -100 \mu C$,and $q_C = +100 \mu C$. The side length is $r = 4 	ext{ m}$.$1$. Force on $C$ due to $A$

Vedclass · Accepted Answer

$5.625 	ext{ N}, 60^{\circ}$

Vedclass · Answer

(A) Let the charges be $q_A = +100 \mu C$,$q_B = -100 \mu C$,and $q_C = +100 \mu C$. The side length is $r = 4 	ext{ m}$.$1$. Force on $C$ due to $A$ $(F_{CA})$: This is a repulsive force directed along the line $AC$ extended away from $A$. The magnitude is $F = \frac{k |q_A q_C|}{r^2} = \frac{9 	imes 10^9 	imes (100 	imes 10^{-6})^2}{4^2} = \frac{9 	imes 10^9 	imes 10^{-8}}{16} = \frac{90}{16} = 5.625 	ext{ N}$.$2$. Force on $C$ due to $B$ $(F_{CB})$: This is an attractive force directed towards $B$. The magnitude is also $F = 5.625 	ext{ N}$ since the charges and distance are the same.$3$. Resultant Force: The angle between $F_{CA}$ and $F_{CB}$ is $120^{\circ}$ (since the interior angle of the equilateral triangle is $60^{\circ}$,the angle between the extension of $AC$ and $CB$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$). The resultant force $F_{	ext{net}} = \sqrt{F^2 + F^2 + 2F^2 \cos(120^{\circ})} = \sqrt{2F^2 + 2F^2(-0.5)} = \sqrt{F^2} = F = 5.625 	ext{ N}$.$4$. Direction: Since the two forces have equal magnitudes,the resultant force bisects the angle between them. The angle between $F_{CA}$ and $F_{CB}$ is $120^{\circ}$. The resultant makes an angle of $60^{\circ}$ with $F_{CA}$ (which is along $AC$). Thus,the angle with $AC$ is $60^{\circ}$.

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