Capacitor with Dielectric — Quick Trick to Find K (JEE Main Concept) ⚡
❓ Question A parallel plate capacitor has a charge of 5 × 10 − 6 C 5 \times 10^{-6}\ \text{C} . A dielectric slab is inserted between the plates and almost fills the space between them. If the induced charge on one face of the slab is 4 × 10 − 6 C 4 \times 10^{-6}\ \text{C} , then find the dielectric constant ( K K ) of the slab. 🖼️ Question Image ✍️ Short Solution Step 1 — Recall the concept When a dielectric slab is inserted into a charged capacitor: The free charge Q Q on the plates remains the same (if disconnected from battery). The dielectric becomes polarized , producing induced charge Q ′ Q' on its surfaces. Relation between induced charge and free charge: Q ′ = Q ( 1 − 1 K ) where K K = dielectric constant. Step 2 — Substitute given values Given: Q = 5 × 10 − 6 C , Q ′ = 4 × 10 − 6 C . So, 4 × 10 − 6 = 5 × 10 − 6 ( 1 − 1 K ) Step 3 — Simplify the equation Cancel 10 − 6 10^{-6} from both s...