MCAT Physics Question 66: Answer and Explanation
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Question: 66
6. Two parallel conducting plates are separated by a distance d. One plate carries a charge +Q and the other carries a charge -Q. The voltage between the plates is 12 V. If a +2 μC charge is released from rest at the positive plate, how much kinetic energy does it have when it reaches the negative plate?
- A. 2.4 × 10–6 J
- B. 4.8 × 10–6 J
- C. 2.4 × 10–5 J
- D. 4.8 × 10–5 J
Correct Answer: C
Explanation:
Recall that the change in potential energy, ΔU, and the change in potential, ΔV, are related by W = ΔU = qΔV. Therefore, ΔU = (2 × 10–6 C) × (–12 V) = –2.4 × 10–5 J. The positive charge is moving from the positive to the negative plate, and is therefore decreasing in potential energy; this is reflected by the fact that the voltage is –12 V rather than +12 V. The potential energy that is lost is converted into kinetic energy, so the charge must gain 2.4 × 10–5 J of kinetic energy.