Consider the following diagram. Two parallel plates are separated in vacuum by 0.200 m. The top plate has a potential of 400 V and the bottom plate has a potential of −600 V. An alpha particle is released from the top plate.The acceleration of the alpha particle between the plates is

Question

BlackTom AI · Accepted Answer

The electric field between the plates is E = ΔV / d. Here ΔV = Vtop − Vbottom = 400 − (−600) = 1000 V and d = 0.200 m, so E = 1000 / 0.200 = 5000 V·m−1 (pointing downward, from the higher potential to the lower).

Relevant quantities: alpha particle charge q = +2e = 2 × 1.602×10−19 C = 3.204×10−19 C. Alpha particle mass m ≈ 4 u = 4 × 1.66054×10−27 kg = 6.64216×10−27 kg. Force on the particle is F = qE and acceleration a = F/m.

Step-by-step calculation:
1) Compute E: E = 1000 V / 0.200 m = 5000 V·m−1.
2) Compute force: F = qE = (3.204×10−19 C)(5000 V·m−1) = 1.602×10−15 N.
   (Intermediate: 3.204×10−19 × 5.0×10^3 = 1.602×10−15.)
3) Compute acceleration: a = F / m = 1.602×10−15 N / 6.64216×10−27 kg
   = (1.602 / 6.64216) × 10^(−15 − (−27)) m·s−2
   = 0.2412 × 10^12 m·s−2 ≈ 0.24 × 10^12 m·s−2.

Why this is correct: the uniform field produces a constant force qE; using the correct alpha charge (2e) and its mass gives the numerical acceleration above. The direction is downward toward the negative plate because the alpha particle is positively charged.

Why the other choices are incorrect:
- 2.39×10^12, 3.20×10^12 and 6.40×10^12 m·s−2 are larger by factors of ~10–25. Such values would result from arithmetic mistakes (wrong ΔV/d, using an incorrect mass by an order of magnitude, or incorrect charge factor). They do not match the correct qE/m calculation.

Conclusion: the calculated acceleration is 0.24 × 10^12 m·s−2, which confirms the provided correct answer.

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