Table of contents

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

25. Electric Potential

Work From Electric Force

Physics

25. Electric Potential

Work From Electric Force

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Multiple Choice

What is the speed of a proton that has been accelerated through $2300 V$ ?

1.2 \times 10^{5} m/s

5.4 \times 10^{5} m/s

7.6 \times 10^{5} m/s

9.3 \times 10^{5} m/s

1.5 \times 10^{6} m/s

6.6 \times 10^{5} m/s

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Verified step by step guidance

Identify the energy gained by the proton due to the potential difference. The energy gained is equal to the charge of the proton multiplied by the potential difference: $ E = qV $, where $ q $ is the charge of the proton $ (1.6 \times 10^{-19} \text{ C}) $ and $ V $ is the potential difference (2300 V).

Calculate the kinetic energy of the proton. Since the proton starts from rest, the kinetic energy gained is equal to the energy gained from the potential difference: $ KE = \frac{1}{2}mv^2 = qV $, where $ m $ is the mass of the proton $ (1.67 \times 10^{-27} \text{ kg}) $.

Rearrange the kinetic energy formula to solve for the speed $ v $: $ v = \sqrt{\frac{2qV}{m}} $.

Substitute the known values into the equation: $ q = 1.6 \times 10^{-19} \text{ C} $, $ V = 2300 \text{ V} $, and $ m = 1.67 \times 10^{-27} \text{ kg} $.

Calculate the speed $ v $ using the rearranged formula to find the speed of the proton after being accelerated through the potential difference.