MCAT Physics Question 84: Answer and Explanation
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Question: 84
9. If the area of a capacitor's plates is doubled while the distance between them is halved, how will the final capacitance (Cf) compare to the original capacitance (Ci)?
- A. Cf = Ci
- B.
- C. Cf = 2Ci
- D. Cf = 4Ci
Correct Answer: D
Explanation:
This question should bring to mind the equation
where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. From this equation, we can infer that doubling the area will double the capacitance, and halving the distance will also double the capacitance. Therefore, the new capacitance is four times larger than the initial capacitance.