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CUET · PHYSICS · PYQ PAPER 2025

A parallel plate air capacitor is charged, and then the battery is disconnected. Now on inserting a dielectric between the plates of the capacitor, which of the following change:
1. potential difference between the plates
2. charge on the plates
3. electric field between the plates
4. energy stored in the capacitor
Choose the correct answer from the options given below:

  1. A (A), (B) and (D) only
  2. B (A), (C) and (D) only
  3. C (B), (C) and (D) only
  4. D (A), (B), (C) and (D)
Verified Solution

Answer & Solution

Correct Answer

(B) (A), (C) and (D) only

Step-by-step Solution

Detailed explanation

Charge Q: Constant (battery disconnected). Capacitance: \(C' = KC\). 1. Potential difference: \(V' = \frac{Q}{C'} = \frac{Q}{KC} = \frac{V}{K}\) (Changes). 2. Charge on the plates: Q (Does not change). 3. Electric field: \(E' = \frac{V'}{d} = \frac{V}{Kd} = \frac{E}{K}\)…
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