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).

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:

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