Question

Two point chargers 9c and 1c ate kept at a distance of 16cm from each other. At what point between these charges should a third charge be placed so that it remains in equilibrium?

Answer 1

Charge to be placed must be negative

REFER the attachment  

let the negative charge introduced has magnitude of q

force b|w q1 and -ve charge must be equal to force b|w q2 and -ve charge

hence

$F_{q_{1-}}=F_{q_{2-}}\\USING\: \:\:\:k\frac{|q_{1}||q_{2}|}{r^{2}} \\\frac{k\times (1\times q)}{(16-x)^{2}} =\frac{k\times (9\times q)}{x^{2}} \\\frac{1}{(16-x)^{2}} =\frac{9}{x^{2}}$

$\frac{x^{2}}{(16-x)^{2}} =\frac{9}{1} \\\frac{x}{16-x}=3\\ x=48-3x\\4x=48\\x=12 cm$

Hence -ve charge must be placed at a distance 12cm from 1c charge