Answer:
x = -7.06 cm
Explanation:
The distance of new charge q from origin = x
We know that force between two charge given as
[tex]F=K\dfrac{q_1q_2}{r^2}[/tex]
Given that the force on the third charge q is zero.
Therefore
[tex]K\dfrac{q\times 3q}{(50-x)^2}=K\dfrac{q\times q}{(40+x)^2}[/tex]
[tex]\dfrac{ 3}{(50-x)^2}=\dfrac{1}{(40+x)^2}[/tex]
3(40+x)² = (50-x)²
[tex]50-x = \sqrt{3}\ (40+x)[/tex]
50 - x = 69.28 + 1.73 x
2.73 x = 50 - 69. 28
x = -7.06 cm
Therefore the third charge on the negative x axis and distance from origin will be 7.06 cm.
Answer:
(x ≈ -7.06)
Explanation:
Given:
∴The total distance between the charges 1 & 2 is 90 units.
For a third particle of charge 'q' to be placed on x axis so that it experiences no force, let us place it at a distance of r units from the charge q.
∴ [tex]F_{qq}=F_{3qq}[/tex]
[tex]k \times \frac{q.q}{r^2} = k \times \frac{q.3q}{(90-r)^2}[/tex]
[tex]r^2+90r-4050=0[/tex]
[tex]r=32.94\ or\ r=-122.94\ (not\ possible)[/tex]
[tex]r\approx32.94[/tex] from the chage q at x= -40
i.e. position of third charge will be at (x ≈ -7.06)
