An electron is placed in an electric field of E = 5.57×10^-11 N/C. What is the acceleration magnitude of the electron (mass m = 9.11×10^-31 kg)?

• A. 9.8 m/s^2
• B. 1.0 m/s^2
• C. 9.8×10^3 m/s^2
• D. 0.98 m/s^2

Answer: (A)

A charged particle in an electric field feels a force F = qE, so its acceleration follows a = F/m = qE/m. For the magnitude, use the absolute charge: a = |q|E/m. An electron has |q| = e ≈ 1.602×10^-19 C. With E = 5.57×10^-11 N/C and m = 9.11×10^-31 kg, the force is F = eE ≈ (1.602×10^-19)(5.57×10^-11) ≈ 8.92×10^-30 N. Then a ≈ F/m ≈ (8.92×10^-30) / (9.11×10^-31) ≈ 9.8 m/s^2. The electron accelerates opposite to the field due to its negative charge, but the magnitude of the acceleration is about 9.8 m/s^2.