Point charges +4q, -q, -q and +4q are kept on the X -axis at point x = 0, x = a and x = 2a respectively. Then (a) only -q is in stable equilibrium (b) all the charges are in stable equilibrium (c) all of the charges are in unstable equilibrium (d) none of the charges is in equilibrium

Answer: (c). Reading the charges as +4q, -q, +4q at x = 0, a, 2a. -q at the middle: equal and opposite pulls from the two +4q, so net force zero. +4q at x = 0: attraction (4kq²)/(a²) towards -q, repulsion (16kq²)/(4a²) = (4kq²)/(a²) away; net zero. Same for the other end. A small shift of any charge makes the nearer attraction grow faster than the balancing force, so it moves further away: all equilibria are unstable. So the answer is (c).

Point charges +4q, -q, -q and +4q are kept on the X -axis at point x… | NEET 1988 (AIPMT)

Q83NEET 1988 · AIPMTMedium

Electric Charge & Coulomb's Law

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

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Answer: (c)

Reading the charges as $+4q, -q, +4q$ at $x = 0, a, 2a$ .
$-q$ at the middle: equal and opposite pulls from the two $+4q$ , so net force zero.
$+4q$ at $x = 0$ : attraction $\dfrac{4kq^2}{a^2}$ towards $-q$ , repulsion $\dfrac{16kq^2}{4a^2} = \dfrac{4kq^2}{a^2}$ away; net zero. Same for the other end.
A small shift of any charge makes the nearer attraction grow faster than the balancing force, so it moves further away: all equilibria are unstable.
So the answer is (c).

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