Question

The point charges of two Q magnitudes are placed at a distance r from each other. A third charge of Q magnitude is placed at the midpoint of the line joining them, what should be its value and nature so that the whole system remains in maximum equilibrium

Answer 1

For equilibrium, net F on each charge = 0 Let identical charges Q be placed at A and B and another charge q is at a distance x from A so that it is in equilibrium.

∴ Force on q due to charge at A in the + X direction

$\frac{1}{4\pi e_{0} } \frac{qp}{ {x}^{2} }$

And force on a due to charge at B in the-X

$\frac{1}{4\pi e_{0} } \frac{qp}{ ({r - x})^{2} }$

For equilibrium, these two forces must be equal i.e.,

$\frac{1}{ {x}^{2} } = \frac{1}{ ({r - x})^{2} } \: or \: x = \frac{r}{2}$

If q was a negative charge, the direction of force due to q at B would be in-X and at A in +X direction.

But, if all the three charges are of same nature, there would be repulsion between charges at A and B also. Hence to have equilibrium among three charges, Q must be opposite of q so that force of attraction between Q.and q=force of repulsion between Q and q.

$\frac{{q}^{2} } {4\pi e_{0} } = \frac{oq}{4\pi e_{0} {r}^{2} } = \frac{qq}{4\pi e_{0} (\frac{r}{2}) ^{2} }$

$= > q = \frac{q}{4}$