A parallel plate capacitor is first charged and then a dielectric s...

Question

A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates (battery disconnected). The quantity that remains unchanged is

Accepted Answer

✅ Correct Answer: (A) Charge

📌 Condition: Battery is disconnected before inserting dielectric.

Quantity | Before Dielectric | After Dielectric (K > 1)

─────────────────────────────────────────────────────────

Charge (Q) | Q | Q → Unchanged ✓

Capacitance (C) | C | KC → Increases

Potential (V) | V | V/K → Decreases

Energy (U) | U | U/K → Decreases

∴ Charge remains unchanged (no path for charge to flow).

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