An electron gun in a television tube uses a uniform electric field to accelerate electrons from rest to 5.0×10^7 m/s over a distance of 1.2 cm. What is the electric field strength?

• A. 5.9×10^3 N/C
• B. 5.9×10^5 N/C
• C. 5.9×10^7 N/C
• D. 9.1×10^5 N/C

Answer: (B)

The accelerating effect of a uniform electric field on a charged particle is a constant force F = qE, giving a = F/m = qE/m. With the electron starting from rest and traveling a distance d, its final speed v satisfies v^2 = u^2 + 2ad, so a = v^2/(2d). Plugging in v = 5.0×10^7 m/s and d = 1.2 cm = 0.012 m gives a ≈ (5.0×10^7)^2 / (2×0.012) ≈ 1.04×10^17 m/s^2. Now relate acceleration to the electric field: E = ma/q. Using electron mass m ≈ 9.11×10^−31 kg and charge magnitude q ≈ 1.60×10^−19 C, E ≈ (9.11×10^−31 kg)(1.04×10^17 m/s^2) / (1.60×10^−19 C) ≈ 5.9×10^5 N/C. So the field strength is about 5.9×10^5 N/C (or V/m).