Match List-I and List-II for a capacitor in following cases.
| List-I | List-II | ||
|---|---|---|---|
| (Reason) | (Effect) | ||
| (A) | Battery disconnected, dielectric inserted between capacitor plates | (I) | Potential difference between plates decreases. |
| (B) | Battery connected, dielectric inserted between capacitor plates | (II) | Charge on plates increases. |
| (C) | Battery disconnected, plate separation increased | (III) | Potential difference between plates remains same, capacitance decreases. |
| (D) | Battery connected, plate separation increased. | (IV) | Capacitance decreases, charge remains same. |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
Use \( C = \varepsilon A/d \) and \( Q = CV \), and ask which quantity is fixed. With the battery disconnected, the charge \( Q \) stays the same. With the battery connected, the potential difference \( V \) stays the same.
(A) Battery disconnected, dielectric inserted between capacitor plates — this matches (I) Potential difference between plates decreases. \( Q \) is fixed and \( C \) becomes \( KC \), so \( V = Q/C \) falls to \( V/K \).
(B) Battery connected, dielectric inserted between capacitor plates — this matches (II) Charge on plates increases. \( V \) is fixed and \( C \) becomes \( KC \), so \( Q = CV \) rises to \( KQ \).
(C) Battery disconnected, plate separation increased — this matches (IV) Capacitance decreases, charge remains same. \( C \propto 1/d \) falls, and with no battery the charge cannot change.
(D) Battery connected, plate separation increased. — this matches (III) Potential difference between plates remains same, capacitance decreases. The battery holds \( V \) fixed while \( C \propto 1/d \) falls.