When 4 C,Q C,and 1 C electrical charges ar | vedclass

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(C) Let the charges be $q_1 = 4 \ C$ at $x = 0$,$q_2 = Q \ C$ at $x = \frac{l}{2}$,and $q_3 = 1 \ C$ at $x = l$. Case $1$: Net force on $4 \ C$ is zer

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$\frac{-1}{4}, -1$

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(C) Let the charges be $q_1 = 4 \ C$ at $x = 0$,$q_2 = Q \ C$ at $x = \frac{l}{2}$,and $q_3 = 1 \ C$ at $x = l$. Case $1$: Net force on $4 \ C$ is zero. The force due to $Q$ and $1 \ C$ on $4 \ C$ must cancel out. $F = k \frac{4 \cdot Q}{(l/2)^2} + k \frac{4 \cdot 1}{l^2} = 0$. $k \frac{4Q}{l^2/4} + \frac{4k}{l^2} = 0 \implies \frac{16Q}{l^2} + \frac{4}{l^2} = 0 \implies 16Q = -4 \implies Q = -\frac{1}{4} \ C$. Case $2$: Net force on $1 \ C$ is zero. The force due to $4 \ C$ and $Q$ on $1 \ C$ must cancel out. $F = k \frac{1 \cdot 4}{l^2} + k \frac{1 \cdot Q}{(l/2)^2} = 0$. $\frac{4k}{l^2} + \frac{kQ}{l^2/4} = 0 \implies \frac{4}{l^2} + \frac{4Q}{l^2} = 0 \implies 4 + 4Q = 0 \implies Q = -1 \ C$. Thus,the values are $-\frac{1}{4} \ C$ and $-1 \ C$.

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