Respuesta :
Solution
In this question we have given
Charge,[tex]q_{1}=-5\times 10^-9C[/tex]
Charge,[tex]q_{2}=-1.5\times 10^-9C[/tex]
Charge,[tex]q_{3} =1.1\times 10^-8C[/tex]
Distance between Charge [tex]q_{1}[/tex] and [tex]q_{2}[/tex],d=43cm=.43m
Let the distance between charge [tex]q_{1}[/tex] and [tex]q_{3} [/tex], is x
Therefore,distance between charge [tex]q_{2}[/tex] and [tex]q_{3} [/tex], will be=(.43-x)
We know By Couloms law, Force on Charge [tex]q_{3}[/tex] due to charge [tex]q_{1}[/tex] which are seperated by distance d is given by following equation
[tex]F_{1}=\frac{k\times q_{1}q_{2}}{d^2}[/tex]............(1)
Here, K=[tex]9\times 10^9 Nm^2C^-2[/tex]
Therefore,
Force on Charge [tex]q_{3}[/tex] due to charge [tex]q_{1}[/tex]
[tex]F_{1}=-\frac{k\times 5\times 10^-9C\times 1.1\times 10^-8C}{x^2}[/tex]..(2)
Similarly,
Force on Charge [tex]q_{3}[/tex] due to charge [tex]q_{2}[/tex]
[tex]F_{2}=-\frac{k\times 1.5\times 10^-9C\times 1.1\times 10^-8C}{(.43-x)^2}[/tex]..(2)
In equilibrium condition,
[tex]F_{1} =F_{2}[/tex]
[tex]-\frac{k\times 5\times 10^-9C\times 1.1\times 10^-8C}{x^2}=-\frac{k\times 1.5\times 10^-9C\times 1.1\times 10^-8C}{(.43-x)^2}[/tex]
[tex]\frac{5\times 10^-9C}{x^2}=\frac{1.5\times 10^-9C}{(.43-x)^2}[/tex]
[tex]5\times 10^-9C \times (.43-x)^2=1.5\times 10^-9C\times (x)^2[/tex]
[tex]3.5x^2-4.3x+.9245=0[/tex]
on solving we get
x=.277m
0r x=28cm
The distance between charge [tex]q_{1}[/tex] and [tex]q_{3} [/tex], is 28cm
