Point charges +4q, -q and +4q are kept on the | vedclass

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(C) First, we calculate the net force on each charge:$1$. For the charge at $x=0$ $(+4q)$: The force due to $-q$ at $x=a$ is attractive, and the force

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All the charges are in unstable equilibrium

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(C) First, we calculate the net force on each charge:$1$. For the charge at $x=0$ $(+4q)$: The force due to $-q$ at $x=a$ is attractive, and the force due to $+4q$ at $x=2a$ is repulsive. The net force is $F = k \frac{(4q)(q)}{a^2} - k \frac{(4q)(4q)}{(2a)^2} = \frac{4kq^2}{a^2} - \frac{4kq^2}{a^2} = 0$.$2$. For the charge at $x=a$ $(-q)$: The force due to $+4q$ at $x=0$ is attractive, and the force due to $+4q$ at $x=2a$ is also attractive. The net force is $F = k \frac{(4q)(q)}{a^2} - k \frac{(4q)(q)}{a^2} = 0$.$3$. For the charge at $x=2a$ $(+4q)$: By symmetry, the net force is $0$.Since the net force on all charges is zero, they are in equilibrium. However, if any charge is displaced slightly, the restoring force does not act to return it to its original position; instead, the net force increases, pushing it further away. Thus, all charges are in unstable equilibrium.

Point charges $+4q, -q$ and $+4q$ are kept on the $x$-axis at points $x = 0, x = a$ and $x = 2a$ respectively. Then:

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