An electron is released from rest in a uniform electric field of 5 × 10⁴ N/C. Its speed after traveling 0.01 m is:
Correct answer: A. 4.2 × 10⁶ m/s
- A. 4.2 × 10⁶ m/s
- B. 8.4 × 10⁶ m/s
- C. 2.1 × 10⁶ m/s
- D. 5.9 × 10⁶ m/s
Explanation
Using energy conservation, work done by the field equals kinetic energy gained: qEΔx = (1/2)mv². For an electron (q = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg), E = 5 × 10⁴ N/C, Δx = 0.01 m, v = √(2qEΔx/m) ≈ 4.2 × 10⁶ m/s.
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