Question

Two charged particles, q1 and q2, are located on the x-axis, with q1 at the origin and q2 initially atx1 = 14.7 mm.In this configuration, q1 exerts a repulsive force of 2.62 µN on q2. Particle q2 is then moved tox2 = 18.0 mm.What is the force (magnitude and direction) that q2 exerts on q1 at this new location? (Give the magnitude in µN.)

Answer 1

Answer:

The force magnitude is 1.75 μN acting outward from the origin towards x₂

Explanation:

The given parameters, are;

The location of q₁ = The origin

The location of q₂ = 14.7 mm from q₁

The repulsive force exerted on q₂ by q₁ = 2.62 μN

The location the particle q₂ is located to =  18.0 mm from q₁

By Coulomb's law, we have;

$F = k\dfrac{q_1 \times q_2 }{r^2}$

Where;

k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)

r = The distance between the particles

F = The force acting between the particles

When r = 14.7 mm F = 2.62 μN

∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸

q₂ × q₁ = 1.48 × 10⁻¹⁸ C²

When the distance is increased to 18.0 mm, we have;

F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N

∴ F = 1.75 μN.

Therefore;

The force magnitude is 1.75 μN outward from the origin towards  x₂.