A proton has a velocity of 1 × 10^5 m/s at point A and a velocity of 3 × 10^5 m/s at point B. If the voltage at point A is 1000 volts, what is the voltage at point B?

• A. 1500 volts
• B. 1000 volts
• C. 2000 volts
• D. 500 volts

Answer: (C)

Verified step by step guidance

1. Identify the known values: the initial velocity \( v_A = 1 \times 10^5 \text{ m/s} \), the final velocity \( v_B = 3 \times 10^5 \text{ m/s} \), and the voltage at point A \( V_A = 1000 \text{ volts} \).
2. Use the principle of energy conservation, which states that the change in kinetic energy is equal to the change in electric potential energy. The equation is \( \Delta KE = q \Delta V \), where \( q \) is the charge of the proton.
3. Calculate the change in kinetic energy: \( \Delta KE = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2 \), where \( m \) is the mass of the proton.
4. Since the charge of a proton \( q = 1.6 \times 10^{-19} \text{ C} \), set the change in kinetic energy equal to \( q \Delta V \) and solve for \( \Delta V \): \( \Delta V = \frac{\Delta KE}{q} \).
5. Determine the voltage at point B using \( V_B = V_A + \Delta V \).