Answer:
Charge Z can be placed at x = -2.7 m or at x = 0.27 m.
Explanation:
The Coulomb force between two charges, [tex]Q_1[/tex] and [tex]Q_2[/tex], separated by a distance, [tex]d[/tex], is given
[tex]F = k\dfrac{Q_1Q_2}{r^2}[/tex]
k is a constant.
For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.
It is to be placed along the x-axis. Hence, it is on the same line as charges X and Y.
Let the charge on Z be Q. It is positive.
Let the distance from charge X be x m. Then the distance from charge Y will be (0.60 - x) m.
Force due to charge X
[tex]F_X = k\dfrac{18Q}{x^2}[/tex]
Force due to charge Y
[tex]F_Y = k\dfrac{-27Q}{(0.60-x)^2}[/tex]
Since both forces are equal and opposite,
[tex]F_X = -F_Y[/tex]
[tex]k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}[/tex]
[tex]\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}[/tex]
[tex]2(0.60-x)^2 = 3x^2[/tex]
[tex]2(0.36-1.20x+x^2) = 3x^2[/tex]
[tex]0.72-2.40x+2x^2 = 3x^2[/tex]
[tex]x^2+2.40x-0.72 = 0[/tex]
Applying the quadratic formula,
[tex]x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}[/tex]
[tex]x = -2.7[/tex] or [tex]x = 0.27[/tex]
Charge Z can be placed at x = -2.7 m or at x = 0.27 m
