Three point charges are placed at the corners | vedclass

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(A) The charges are $q_A = +100 \ \mu C$,$q_B = +100 \ \mu C$,and $q_C = -100 \ \mu C$. The side length $r = 10 \ cm = 0.1 \ m$.The force on $B$ due t

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$9 	imes 10^3 \ N$

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(A) The charges are $q_A = +100 \ \mu C$,$q_B = +100 \ \mu C$,and $q_C = -100 \ \mu C$. The side length $r = 10 \ cm = 0.1 \ m$.The force on $B$ due to $A$ $(F_A)$ is repulsive and acts along the line $AB$ away from $A$: $F_A = \frac{k |q_A q_B|}{r^2} = \frac{9 	imes 10^9 	imes (100 	imes 10^{-6}) 	imes (100 	imes 10^{-6})}{(0.1)^2} = 9 	imes 10^3 \ N$.The force on $B$ due to $C$ $(F_C)$ is attractive and acts along the line $BC$ towards $C$: $F_C = \frac{k |q_C q_B|}{r^2} = \frac{9 	imes 10^9 	imes (100 	imes 10^{-6}) 	imes (100 	imes 10^{-6})}{(0.1)^2} = 9 	imes 10^3 \ N$.The angle between $F_A$ and $F_C$ is $120^\circ$ because the angle between the lines $AB$ and $BC$ is $60^\circ$,and the force $F_C$ is directed towards $C$ (making an angle of $180^\circ - 60^\circ = 120^\circ$ with $F_A$).The resultant force $F_{net} = \sqrt{F_A^2 + F_C^2 + 2 F_A F_C \cos(120^\circ)}$.Since $F_A = F_C = F = 9 	imes 10^3 \ N$,$F_{net} = \sqrt{F^2 + F^2 + 2F^2(-0.5)} = \sqrt{F^2} = F = 9 	imes 10^3 \ N$.

Three point charges are placed at the corners of an equilateral triangle with sides of $10 \ cm$ as shown in the figure. The resultant force on the charge at $B$ will be:

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