Question

A particle with charge 2-µC is placed at the origin, an identical particle, with the same charge,
is placed 2m from the origin on the x axis, and a third identical particle, with the same charge,
is placed 2m from the origin on the y axis. The magnitude of the force on the particle at the
origin is:
A. 9.0 × 10-3 N
B. 6.4 × 10-3 N
C. 1.3 × 10-2 N
D. 1.8 × 10-2 N
E. 3.6 × 10-2 N

Answer 1

Answer:

1.3 × 10-2 N

Explanation:

Coulomb's law relate the force between two charges as

F = kq₁q₂ / d² where q₁ is at the origin

q₁ = q₂ = q₃= 2 µC = 2 × 10⁻⁶ C

F₁₂ = 8.99 × 10⁹ × 2 × 10⁻⁶ × 2 × 10⁻⁶ / 4 = 0.009 N along x axis

F₁₃ along y axis = 8.99 × 10⁹× 2 × 10⁻⁶ × 2 × 10⁻⁶ / 4 = 0.009 N along y axis

the magnitude of the force on the particle at the origin = √ (0.009² + 0.009²) = 1.27 × 10 ⁻² N approx 1.3 × 10-2 N

Answer 2

The   magnitude of the force on the particle at the origin is 1.3 × 10-2 N

Coulomb's law:

This law related to the force between two charges likewise

F = kq₁q₂ / d²

here q₁ is at the origin

Now

q₁ = q₂ = q₃= 2 µC = 2 × 10⁻⁶ C

Now

F₁₂ = 8.99 × 10⁹ × 2 × 10⁻⁶ × 2 × 10⁻⁶ / 4 = 0.009 N along x axis

Now

F₁₃ along y axis = 8.99 × 10⁹× 2 × 10⁻⁶ × 2 × 10⁻⁶ / 4 = 0.009 N along y axis

So, the magnitude is

= √ (0.009² + 0.009²)

= 1.27 × 10 ⁻² N

= 1.3 × 10-2 N

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