Capacitance is C = Q/V, so V = Q/C. With Q=28 C and C=12 mF, V=28/0.012≈2333 V=2.33 kV.
Formula / Key Relation
C = Q/V
V = Q/C
= 28/(12×10⁻³)
≈ 2.333×10³ V
= 2.33 kV
Detailed Explanation
Concept & ReasoningThe defining relationship for an isolated capacitor/conductor system is Q = CV. Convert millifarads to farads before division: 12 mF = 0.012 F. Then V = 28/0.012 = 2333.3 V. The source option typography may show an incorrect unit in extracted text; the physical quantity asked is potential, so the unit must be volts, here approximately 2.33 kV.How to Solve It in the ExammF → 10⁻³ F before using V=Q/C.Important Exam PointCheck the dimensional unit of the answer; electric potential must be in volts.Related Revision PathThis question sits in the revision path Electromagnetic Fields → Electrostatics → Capacitance…
Concept & Reasoning
The defining relationship for an isolated capacitor/conductor system is Q = CV. Convert millifarads to farads before division: 12 mF = 0.012 F. Then V = 28/0.012 = 2333.3 V. The source option typography may show an incorrect unit in extracted text; the physical quantity asked is potential, so the unit must be volts, here approximately 2.33 kV.
How to Solve It in the Exam
mF → 10⁻³ F before using V=Q/C.
Important Exam Point
Check the dimensional unit of the answer; electric potential must be in volts.
Related Revision Path
This question sits in the revision path Electromagnetic Fields → Electrostatics → Capacitance .
Quick Trick
mF → 10⁻³ F before using V=Q/C.
Common Mistake
Reading a corrupted 'kV' as 'kΩ' is an OCR/extraction error, not a physics result.
Exam Tip
Check the dimensional unit of the answer; electric potential must be in volts.
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