Three charges are placed along the x-axis: q1 = 8.0 x 10⁻⁶ C at x=0, q2 = 2.0 x 10⁻⁶ C at x=2.0 m, and q3 = 5.0 x 10⁻⁶ C at x=4.0 m. Calculate the net force on q2.

To find the net force on the $2.0 \times 10^{-6}$ C charge, we will use Coulomb's Law and the principle of superposition.

Let the charges be:
$q_1 = 8.0 \times 10^{-6}$ C (left charge)
$q_2 = 2.0 \times 10^{-6}$ C (middle charge)
$q_3 = 5.0 \times 10^{-6}$ C (right charge)

The distances are:
$r_{12} = 2.0$ m (distance between $q_1$ and $q_2$)
$r_{23} = 2.0$ m (distance between $q_2$ and $q_3$)

Coulomb's constant $k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$.

Since all charges are positive, the forces are repulsive.

• The force $F_{12}$ (on $q_2$ due to $q_1$) pushes $q_2$ to the right.
• The force $F_{32}$ (on $q_2$ due to $q_3$) pushes $q_2$ to the left.

Step 1: Calculate the magnitude of the force $F_{12}$ exerted by $q_1$ on $q_2$.

$$F_{12} = \frac{k |q_1 q_2|}{r_{12}^2}$$ $$F_{12} = \frac{(9 \times 10^9 \, \text{N m}^2/\text{C}^2) (8.0 \times 10^{-6} \, \text{C}) (2.0 \times 10^{-6} \, \text{C})}{(2.0 \, \text{m})^2}$$ $$F_{12} = \frac{9 \times 8.0 \times 2.0 \times 10^{9-6-6}}{4.0} \, \text{N}$$ $$F_{12} = \frac{144 \times 10^{-3}}{4.0} \, \text{N}$$ $$F_{12} = 36 \times 10^{-3} \, \text{N}$$

Step 2: Calculate the magnitude of the force $F_{32}$ exerted by $q_3$ on $q_2$.

$$F_{32} = \frac{k |q_3 q_2|}{r_{23}^2}$$ $$F_{32} = \frac{(9 \times 10^9 \, \text{N m}^2/\text{C}^2) (5.0 \times 10^{-6} \, \text{C}) (2.0 \times 10^{-6} \, \text{C})}{(2.0 \, \text{m})^2}$$ $$F_{32} = \frac{9 \times 5.0 \times 2.0 \times 10^{9-6-6}}{4.0} \, \text{N}$$ $$F_{32} = \frac{90 \times 10^{-3}}{4.0} \, \text{N}$$ $$$$