TP 19.16

Given:

q₁ = +12.0 x 10^-6 C

q₂ = -24.0 x 10^-6 C

q₃ = +36.0 x 10^-6 C

d = 0.160 m

Find: x, where x is the distance from charge q₁ to charge q₂. 

Force that charge 1 exerts on charge 2:

F₁₂ = k |q₁||q₂|/x²

Note, the direction of this force is to the left.

Force that charge 3 exerts on charge 2:

F₃₂ = k |q₃||q₂|/(2d - x)²

Note, the direction of this force is to the right.

Net force that acts on charge 2:

F₂ = F₁₂ + F₃₂= - k |q₁||q₂|/x² + k |q₃||q₂|/(2d - x)² = 0

- k |q₁||q₂|/x² + k |q₃||q₂|/(2d - x)² = 0

-  |q₁|/x² +  |q₃|/(2d - x)² = 0

  |q₁|/x² =  |q₃|/(2d - x)² 

q/x² = 3q/(2d - x)² 

1/x² = 3/(2d - x)² 

(2d - x)² = 3 x²

4d² - 4dx + x² = 3x²

2x² + 4dx - 4d² = 0

x² + 2dx - 2d² = 0

x = - d +/- (d² + 2 d²)^1/2

x = - d + (3)^1/2 d  = 0.732 d = 0.732 (0.160 m) = 0.117 m