What electric field strength is needed to balance the weight of an electron (mass 9.11×10^-31 kg, charge -e = -1.60×10^-19 C)?

• A. 5.57×10^-11 N/C directed downward
• B. 5.57×10^-11 N/C directed upward
• C. 1.02×10^-7 N/C directed upward
• D. 1.02×10^-7 N/C directed downward

Answer: (A)

Balancing the forces on the electron. The weight pulls downward with magnitude mg, while the electric force is Fe = qE. For a negative charge, the electric force points opposite the electric field, so to push the electron upward (to balance the downward weight) the field must point downward. The magnitudes must match: |q|E = mg, so E = mg/|q|.

Plug in m = 9.11×10^-31 kg, g ≈ 9.8 m/s^2, and |q| = 1.60×10^-19 C:

mg ≈ 9.11×10^-31 × 9.8 ≈ 8.93×10^-30 N.

E ≈ (8.93×10^-30 N) / (1.60×10^-19 C) ≈ 5.6×10^-11 N/C.

Direction: since the electron’s charge is negative, a downward field produces an upward electric force, which balances the downward weight. So the required field strength is about 5.57×10^-11 N/C directed downward.