Question

2. Two charges 160μC and -250μC are placed together at a distance of 150 centimeter apart. Find the force of attraction between each other. Also explain the nature of the force [Repulsing force = +ve, Attraction force = -ve]
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Answer 1

The answer is -0.016 N, and as it negative, the nature of the force is attractive.

Finding the force of attraction

Coulomb's Law is given by :

$\boxed {F = \frac{kq_{1}q_{2}}{r^{2}}}$

Solving

• F = 9 × 10⁹ × 160 × 10⁻⁶ × -250 × 10⁻⁶ / (150 × 10⁻³)²
• F = -360000 × 10⁻³ / 22500 × 10⁻⁶
• F = -16 × 10⁻³
• F = -0.016 N