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

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(B) The charges are $q_A = 1 \mu C$,$q_B = -1 \mu C$,and $q_C = 2 \mu C$. The side length $r = 10 \ cm = 0.1 \ m$.The force exerted by charge at $A$ o

Vedclass · Accepted Answer

$1.8$

Vedclass · Answer

(B) The charges are $q_A = 1 \mu C$,$q_B = -1 \mu C$,and $q_C = 2 \mu C$. The side length $r = 10 \ cm = 0.1 \ m$.The force exerted by charge at $A$ on $C$ is $F_{AC} = \frac{k |q_A q_C|}{r^2} = \frac{9 	imes 10^9 	imes 1 	imes 10^{-6} 	imes 2 	imes 10^{-6}}{(0.1)^2} = 1.8 \ N$ (repulsive,along $AC$ extended).The force exerted by charge at $B$ on $C$ is $F_{BC} = \frac{k |q_B q_C|}{r^2} = \frac{9 	imes 10^9 	imes 1 	imes 10^{-6} 	imes 2 	imes 10^{-6}}{(0.1)^2} = 1.8 \ N$ (attractive,towards $B$).The angle between these two force vectors is $120^\circ$ because the interior angle of the equilateral triangle is $60^\circ$.The resultant force $F_{net} = \sqrt{F_{AC}^2 + F_{BC}^2 + 2 F_{AC} F_{BC} \cos(120^\circ)}$.Since $F_{AC} = F_{BC} = F = 1.8 \ N$,$F_{net} = \sqrt{F^2 + F^2 + 2F^2(-0.5)} = \sqrt{F^2 + F^2 - F^2} = F$.Therefore,$F_{net} = 1.8 \ N$.

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: (in $N$)

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