Capacitance from charge and voltage: A capacitor has 25 V across its plates and stores a charge of 1500 µC. Compute the capacitance C (express your answer in appropriate SI units).

Difficulty: Easy

Correct Answer: 60 µF

Explanation:


Introduction:
Capacitance relates stored charge to voltage. For a lumped capacitor, the basic definition C = Q / V directly gives the required value when charge and voltage are known. Unit handling is crucial to avoid order-of-magnitude mistakes.


Given Data / Assumptions:

  • Voltage V = 25 V
  • Charge Q = 1500 µC
  • Ideal capacitor with linear behavior


Concept / Approach:

Use the definition C = Q / V. Convert microcoulombs to coulombs: 1 µC = 1e-6 C. After calculation, express the result in a convenient unit (µF).


Step-by-Step Solution:

Convert Q: Q = 1500 µC = 1500 * 1e-6 C = 0.0015 CApply C = Q / VC = 0.0015 / 25 = 0.00006 FConvert to µF: 0.00006 F = 60 µF


Verification / Alternative check:

Back-calculate charge with C = 60 µF at 25 V: Q = C * V = 60e-6 * 25 = 0.0015 C = 1500 µC, confirming correctness.


Why Other Options Are Wrong:

  • 60 pF: Off by 10^6 due to unit confusion.
  • 16.67 mF and 37.5 mF: Overestimates; would imply unrealistically large energy for given Q and V.
  • 6.0 µF: Off by a factor of 10.


Common Pitfalls:

  • Forgetting to convert µC to C before dividing.
  • Confusing µF with mF and pF.


Final Answer:

60 µF

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