A proton with an initial speed of 7.0×10^5 m/s is brought to rest by an electric field. What is the stopping potential for the proton?

• A. 2557 V
• B. -2557 V
• C. 25.57 V
• D. 0 V

Answer: (A)

Stopping potential represents the electric potential energy needed to wipe out the particle’s kinetic energy. For a charged particle, the work done by moving through a potential difference V is qV, so the stopping condition is KE = qV. Since the proton has charge e, the stopping potential is V = KE/e.

Compute the kinetic energy with nonrelativistic formula: KE = (1/2) m v^2. With m = proton mass = 1.6726×10^-27 kg and v = 7.0×10^5 m/s, KE ≈ 0.5 × 1.6726×10^-27 × (7.0×10^5)^2 ≈ 4.10×10^-16 J.

Divide by the proton charge e = 1.602×10^-19 C to get the stopping potential: V ≈ (4.10×10^-16 J) / (1.602×10^-19 C) ≈ 2.56×10^3 V.

So the stopping potential is about 2.56 kV (2.56×10^3 V). The positive magnitude corresponds to the required potential difference to stop the positively charged proton.