Question

Particles q1, q2, and q3 are in a straight line. Particles q1 = -5.00 x 10^-6 C, q2 = +2.50 x 10^-6 C, and q3 = -2.50 x 10^-6 C. Particles q1 and q2 are separated by 0.500 m. Particles q2 and q3 are separated by 0.250 m. What is the net force on q2

Answer 1

When q₂ is place between q₁ and q₃, the net force on q₂ is 0.486N.

What is the electrostatic force?

The electrostatic force of attraction or repulsion is proportional to the product of magnitude of charges and inversely proportional to the square of distance between them.

Force on q₂ due to q₁

F₁₂ = kq₁q₂ / r₁₂²

Substitute the values provided , we get

F₁₂ = 9 x 10⁹ x 5 x 10⁻⁶ x 2.5 x 10⁻⁶ / (0.5)²

F₁₂ = 0.414 N

Force on q₂ due to q₃ placed at distance 0.25m

F₂₃ =kq₂q₃ / r₂₃²

Substitute the values, we get

F₂₃ =  9 x 10⁹ x 2.5 x 10⁻⁶ x 2.5 x 10⁻⁶ / (0.25)²

F₂₃ = 0.9N

The net force F =F₂₃ -F₁₂

F =0.9 - 0.414 = 0.486 N

Thus, the net force of q₂ is 0.486 N.

Learn more about electrostatic force.

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