Answer:
0.5 m
Explanation:
Two charges each of magnitude q
Let the third charge is Q is placed at a distance x from the origin so that the charge is in equilibrium.
The force on Q due to q at origin is balanced by the charge on Q due to the charge q placed at x = 1 m.
So,
[tex]\frac{KQq}{x^{2}}=\frac{KQq}{\left ( 1-x \right )^{2}}[/tex]
1 - x = x
1 = 2x
x = 0.5 m
Thus, the third charge is placed at x = 0.5 m .