Physics Alternating Current PMDC Conceptual Practice
PMDC Verified Question 53 of 127
When an alternating voltage \(v = V_0 \sin(\omega t)\) is applied to a pure capacitor, what is the phase relationship between the applied voltage \(v\) and the instantaneous charge \(q\) on the capacitor plates?
A
Voltage and charge are in phase (\(\phi = 0^\circ\))
B
Voltage leads charge by \(90^\circ\)
C
Charge leads voltage by \(90^\circ\)
D
Voltage and charge are \(180^\circ\) out of phase
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: Voltage and charge are in phase (\(\phi = 0^\circ\))
Concept:

The fundamental relation between charge on capacitor plates and potential difference across them is \(q(t) = C v(t)\).

Formula / Reaction:

$$q(t) = C V_0 \sin(\omega t) = Q_0 \sin(\omega t)$$

Solution:

  • Because capacitance \(C\) is a positive constant scalar, \(q(t)\) is directly proportional to \(v(t)\) at every instant.


  • Therefore, charge \(q\) and voltage \(v\) are exactly in phase (\(\phi = 0^\circ\)).


  • (Note: Current \(i = dq/dt\) leads both charge and voltage by \(90^\circ\)).


Why other options are incorrect:

  • Opt_B, Opt_C, Opt_D: There is no phase difference between charge and voltage in a capacitor.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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