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ExamsJEE AdvancedPhysics

A parallel-plate capacitor is charged by connecting it to a 2 V battery. The battery is then disconnected, and a glass slab (dielectric) is inserted between the plates. Which of the following pairs of physical quantities both decrease?

  1. Energy stored and capacitance
  2. Charge and potential difference
  3. Capacitance and charge
  4. Potential difference and energy stored

Correct answer: Potential difference and energy stored

Solution

When the battery is disconnected before inserting the dielectric, charge Q is conserved (no path for charge to flow). Inserting a glass slab (dielectric constant k > 1) increases capacitance: C' = k*C. Since Q is constant: V' = Q/C' = Q/(kC) = V/k — the potential difference decreases. Energy stored: U' = Q²/(2C') = Q²/(2kC) = U/k — the energy stored also decreases. Capacitance increases, not decreases. Therefore, the pair that both decrease is 'Potential difference and energy stored'.

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