Two force point charges $$+q$$ and $$+4q$$ are a distance $$\ell$$ apart. A third charge $$q_1$$ is so placed that the entire system is in equilibrium. Find the location, magnitude and sign of the third charge.
To get zero net force on placed charge it must be negative.
On balancing the forces,
$$\cfrac{Kqq'}{(l-x)^{2}} = \cfrac{Kq' (4q)}{x^{2}}$$
$$x^{2} = 4 (l-x)^{2}$$
$$x^{2} = 4l^{2} + 4x^{2} -8lx $$
$$3x^{2} -8lx +4l^{2} = 0 $$
$$x= \cfrac{8l \pm \sqrt{64l^{2}-48l^{2}}}{6} = \cfrac{8l \pm 4l}{6} = 2l$$ and $$\cfrac{2l}{3}$$
$$x=\cfrac{l}{3}$$ is the suitable answer.
Now, on balancing the forces on charge 4q
$$\cfrac{K(q')4q}{\left( \cfrac{2l}{3} \right)^{2} } = \cfrac{K(q)(4q)}{l^{2}}$$
$$q' = - \cfrac{4q}{9}$$
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