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⚛️ Physics  ·  Electrostatic Potential and Capacitance  ·  NEET & JEE

Two capacitors of 3 µF and 6 µF are joined in series across 9 V. The potential difference across the 3 µF capacitor is:

Answer: 6 V.

  • A 3 V
  • B 4.5 V
  • C 9 V
  • D 6 V

Correct answer: D. 6 V

Explanation: Series capacitance = (3×6)/(3+6) = 2 µF, so Q = 2 µF × 9 V = 18 µC. Across the 3 µF: V = Q/C = 18/3 = 6 V.

Radial electric field lines and concentric red equipotential lines around a point charge (an electron), with field lines pointing inward

Field lines (radial) and equipotential lines (concentric circles) of a point charge. Equipotential surfaces are always perpendicular to the field lines. Image: Sjlegg, Public Domain, via Wikimedia Commons.

Concept context

Electric potential, equipotential surfaces, potential energy of charge systems, conductors, dielectrics, capacitors, combinations, and energy storage.

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