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Calculate the potential of a metal plate of charge 28C and capacitance 12 mF.

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Answer & Explanation
Correct AnswerB. 2.33 k ohm

Quick Explanation

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…

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Previous Exam Appearances

Assistant Engineer (Electrical), Class-2, Road and Building Department2025Q. 148
Assistant Engineer (Electrical), Class-2, Road and Building Department2025Q. 148
Assistant Engineer (Electrical), Class-2, Road and Building Department2025Q. 148
Assistant Engineer (Electrical), Class-2, Road and Building Department2025Q. 148
Assistant Engineer (Electrical), Class-2, Road and Building Department2025Q. 148
Assistant Engineer (Electrical), Class-2, Road and Building Department2025Q. 148
Assistant Engineer (Electrical), Class-2, Road and Building Department2025Q. 148
Assistant Engineer (Electrical), Class-2, Road and Building Department2025Q. 148
Assistant Engineer (Electrical), Class-2, Road and Building Department2025Q. 148

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