Answer:
0.245 m
Explanation:
We are given that
[tex]q_1=-5nC=-5\times 10^{-9} C[/tex]
Where [tex]1nC=10^{-9} C[/tex]
[tex]q_2=-2.00nC=-2\times 10^{-9}C[/tex]
Distance between q1 and q2=0.40 m
We have to find the equilibrium for a positive charge placed between them.
Let q be the positive charge placed at distance x from q1 .
r1=x
r2=0.40-x
At equilibrium
Total force =0
[tex]\frac{kq\times 5\times 10^{-9}}{x^2}-\frac{kq\times 2\times 10^{-9}}{(0.40-x)^2}=0[/tex]
Where electric force=[tex]\frac{kq_1q_2}{r^2}[/tex]
[tex]\frac{5\times 10^{-9}\times kq}{x^2}=\frac{2\times 10^{-9}\times kq}{(0.40-x)^2}[/tex]
[tex]5(0.40-x)^2=2x^2[/tex]
[tex]0.8-4x+5x^2=2x^2[/tex]
[tex]5x^2-2x^2-4x+0.8=0[/tex]
[tex]3x^2-4x+0.8=0[/tex]
[tex]x=\frac{4\pm\sqrt{16-4\times 3\times 0.8}}{3(2)}[/tex]
By using the formula
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{4\pm 2.5298}{6}[/tex]
By solving we get
[tex]x=1.0883,0.245[/tex]
[tex]x=1.0883>0.40 [/tex]
It is not possible
Therefore, the equilibrium point for a positive charge placed between them is given by
x=0.245 m
