Question

Two free point charges +q and +4q are at the co-ordinates (0,0) and (a, 0). A third charge is placed on the line joining them so that the entire system is in equilibrium. The value and position of the third charge is

- \frac{4 q}{9} a t (\frac{a}{3}, 0)

\frac{2 q}{3} a t (\frac{a}{3}, 0)

- \frac{2 q}{3} a t (\frac{a}{2}, 0)

\frac{4 q}{9} a t (\frac{2 a}{3}, 0)

Solution

The correct option is A $- \frac{4 q}{9} a t (\frac{a}{3}, 0)$
$x = \frac{d}{\sqrt{\frac{q_{2}}{q_{1}}} + 1} = \frac{a}{\sqrt{\frac{4 q}{q}} + 1} = \frac{a}{3} (y = 0)$
$\to F_{12} + \to F_{13} = 0$
$\frac{1}{4 π ϵ_{0}} \frac{q Q}{(\frac{a^{2}}{9})} = - \frac{1}{4 π ϵ_{0}} \frac{q \times 4 q}{a^{2}}$
$Q = - \frac{4 q}{9}$

Physics

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