Question

Two charges 2*10^-6 and 1*10^-6 are placed at a separation of 10 cm. Where should a third charge be placed such that it experiences no net force due to the other two charges?

Answer 1

Answer:

Explanation:

$Q_{1} = 1(10^{-6})$

$Q_{2} = 2(10^{-6})$

r = 10cm =>0.1m

let us keep the distance from charge 2 to the third charge as 'x'.

$F_{32} = K \frac{2(10^-6)q}{x^{2}}$               $K = 9(10^{9})$

$F_{32} = \frac{1.8(10^4)q}{x^{2}}$

$F_{31} = K \frac{10^{-6}q}{(10-x)^{2}}$

$F_{31} = \frac{9(10^{3})q}{(10-x)^{2}}$

as it has to be placed where there'll be no net charge it has to be in equilibrium.

so, $F_{31}=F_{32}$

$\frac{9(10^{3})q}{(10-x)^{2}}= \frac{1.8(10^{4})q}{x^{2}}$

$\frac{x^{2}}{(10-x)^{2}}=  \frac{1.8(10)^{4}}{9(10^{3})}$

$x^{2} = 2(10-x)^{2}$

$x^{2}-40x+200=0$

$x = 20+10\sqrt{2} , 20-10\sqrt{2}$

hence the third charge can be placed in x postion from charge 1