Basic capacitor relationship: Given electric charge Q = 60 µC on the plates and voltage V = 12 V across a capacitor, compute the capacitance value.

Introduction / Context:
This problem checks the fundamental capacitor equation linking charge (Q), voltage (V), and capacitance (C). Such conversions are common in electronics when sizing capacitors for filters or timing networks.

Given Data / Assumptions:

• Charge Q = 60 µC.
• Voltage V = 12 V.
• Ideal capacitor; no leakage or series resistance considered.

Concept / Approach:
The basic relationship is Q = C * V. Solving for C gives C = Q / V. Keep units consistent by converting microcoulombs to coulombs where needed: 1 µC = 10^-6 C.

Step-by-Step Solution:
Q = 60 µC = 60 * 10^-6 CV = 12 VC = Q / VC = (60 * 10^-6) / 12C = 5 * 10^-6 F = 5 µF

Verification / Alternative check:
Reverse the calculation: Q' = C * V = 5 µF * 12 V = 60 µC, which matches the given charge. This confirms the computed capacitance is correct.

Why Other Options Are Wrong:

• 720 µF: Exaggerated by a factor of 144; does not satisfy Q = C * V.
• 50 µF: Ten times too large; would imply Q = 600 µC at 12 V.
• 12 µF: More than double the correct value; would imply Q = 144 µC.

Common Pitfalls:
Forgetting to convert microcoulombs to coulombs, or misplacing decimal points during division, commonly leads to errors of 10x or 100x. Always keep track of powers of ten when working with micro-units.

Final Answer:
5 µF