Four electric charges 2 mu C, Q, 4 mu C and 1 | vedclass

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Question

(B) The force between two point charges is given by Coulomb's Law: $F = k \frac{q_1 q_2}{r^2}$.Let the charges be $q_0 = 2 \mu C$ at $x=0$,$q_1 = Q$ a

Vedclass · Accepted Answer

$-1.75$

Vedclass · Answer

(B) The force between two point charges is given by Coulomb's Law: $F = k \frac{q_1 q_2}{r^2}$.Let the charges be $q_0 = 2 \mu C$ at $x=0$,$q_1 = Q$ at $x=1 \ cm$,$q_2 = 4 \mu C$ at $x=2 \ cm$,and $q_3 = 12 \mu C$ at $x=4 \ cm$.The net force on the charge at the origin $(q_0)$ is the sum of forces exerted by $q_1, q_2,$ and $q_3$.$F_{net} = k q_0 \left( \frac{Q}{(1 	imes 10^{-2})^2} + \frac{4 	imes 10^{-6}}{(2 	imes 10^{-2})^2} + \frac{12 	imes 10^{-6}}{(4 	imes 10^{-2})^2} ight) = 0$.Since $k q_0 
eq 0$,we have: $\frac{Q}{10^{-4}} + \frac{4 	imes 10^{-6}}{4 	imes 10^{-4}} + \frac{12 	imes 10^{-6}}{16 	imes 10^{-4}} = 0$.$10^4 Q + 10^{-2} + 0.75 	imes 10^{-2} = 0$.$10^4 Q + 1.75 	imes 10^{-2} = 0$.$Q = -1.75 	imes 10^{-6} \ C = -1.75 \mu C$.

Four electric charges $2 \mu C, Q, 4 \mu C$ and $12 \mu C$ are placed on the $x$-axis at distances $x=0, 1 \ cm, 2 \ cm$ and $4 \ cm$ respectively. If the net force acting on the charge at the origin is zero,then $Q=$ (in $\mu C$)

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