Answer:
Two possible points
x= 0.67 cm to the right of q1
x= 2 cm to the left of q1
Explanation:
Electrostatic Forces
If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude
[tex]\displaystyle f=k\frac{q_1\ q_2}{d^2}[/tex]
We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.
[tex]\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}[/tex]
[tex]\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}[/tex]
[tex]\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}[/tex]
[tex]\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}[/tex]
[tex]\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}[/tex]
Equating
[tex]\displaystyle F_{13}=F_{23}[/tex]
[tex]\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}[/tex]
Operating and simplifying
[tex]\displaystyle (0.02-x)^2=4x^2[/tex]
To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.
[tex]\displaystyle 0.02-x=\pm 2x[/tex]
Assuming the positive sign :
[tex]\displaystyle 0.02-x= 2x[/tex]
[tex]\displaystyle 3x=0.02[/tex]
[tex]\displaystyle x=0.00667\ m[/tex]
[tex]x=0.67\ cm[/tex]
Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative
[tex]\displaystyle 0.02-x=-2x[/tex]
[tex]\displaystyle x=-0.02\ m[/tex]
[tex]\displaystyle x=-2\ cm[/tex]
The negative sign of x means q3 is located to the left of q1 (assumed in the origin).
