A particle of mass 1.42 g is placed in a uniform downward electric field of 700 N/C. What is the charge (sign and magnitude) required for it to remain stationary?

• A. -1.58×10^-5 C
• B. -1.99×10^-5 C
• C. -7.00×10^-6 C
• D. +1.99×10^-5 C

Answer: (B)

To sit still, the electric force must balance the weight of the particle. The weight is mg downward, with m = 1.42 g = 1.42×10^-3 kg and g ≈ 9.8 m/s^2, so mg ≈ 1.42×10^-3 × 9.8 ≈ 1.3916×10^-2 N. The electric force is F_E = qE, and the field is downward. For the electric force to point upward (to oppose gravity), q must be negative. Setting the magnitudes equal for equilibrium gives |q|E = mg, so |q| = mg / E = (1.3916×10^-2 N) / (700 N/C) ≈ 1.99×10^-5 C. Since the force must be upward, the charge is negative: q ≈ -1.99×10^-5 C.