Hi there!
1)
Since the charge placed in the middle is positive, we know that the particle is being repelled.
The particle experiences a greater repelling force by the 9μC charge. We can use the equation for electric force:
[tex]F_E = \frac{kq_1q_2}{r^2}[/tex]
k = Coulomb's Constant
q₁, q₂: Charges (C)
r = distance between charges (m)
This is a VECTOR quantity, so we must subtract the forces since they point in opposite directions.
Force from 9μC particle:
[tex]F_E = \frac{k(.000005)(.000009)}{2^2} = .101\\\\[/tex]
Force from 4μC particle:
[tex]F_E = \frac{k(0.000005)(0.000004)}{2^2} = 0.0450[/tex]
Subtract:
[tex].101 - 0.0450 = \boxed{0.561 N}[/tex]
2)
We can find a position by setting the two equations equal to one another. (Both repelling forces must be EQUAL for the force = 0 N)
Let the distance between the 9μC and 5μC charge equal 'x', and the distance between the 4μC and 5μC charge equal '4 - x'.
[tex]\frac{k(0.000009)(0.000005)}{x^2} = \frac{k(0.000004)(0.000005)}{(4 - x)^2}[/tex]
Cancel out 'k' and the 5μC value.
[tex]\frac{0.000009}{x^2} = \frac{0.000004}{(4 - x)^2}[/tex]
Solve for 'x' using a graphing utility.
[tex]\boxed{x = 2.4 m}[/tex]
