# Dielectrics and Capacitance > Dielectrics and capacitance for NEET Physics: parallel plate capacitor, dielectric constant K, series and parallel combinations, stored energy. - Canonical URL: https://prepelephant.com/topics/neet-ug/physics/dielectrics-and-capacitance - Exam / course: NEET-UG · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Dielectrics and Capacitance", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/dielectrics-and-capacitance ## Direct answer A capacitor stores charge at the cost of voltage: for the parallel-plate geometry C = ε0 A/d, and inserting a dielectric slab of constant K multiplies the capacitance to C = K ε0 A/d by reducing the internal field to E/K through induced polarisation. The stored energy is U = (1/2) C V^2 = Q^2/2C = (1/2) Q V. Series capacitors add reciprocals (equal charge); parallel ones add directly (equal voltage). The sharpest NEET theme is the slab's insertion — with the battery disconnected the charge is frozen while voltage and energy fall by K; connected, the voltage is pinned while charge and energy rise by K; every arrow follows from which quantity is conserved. ## What you must remember - **Base formula:** C = ε0 A/d for plates in vacuum or air; inserting a dielectric of constant K gives C = K C0, with K taken as 1 for air, roughly 2-6 for oils, paper and mica, about 80 for water and over 100 for certain ceramics. - **Dielectric action:** polar molecules align and non-polar molecules acquire induced dipole moments; the induced surface charge opposes the applied field, reducing E to E0/K and hence the voltage for the same charge. - **Energy trio:** U = (1/2) C V^2 = Q^2/(2C) = (1/2) Q V; pick the form whose variables stay known. - **Series:** 1/C_eq = 1/C1 + 1/C2; the equivalent is smaller than the smallest member; charges equal, voltages share inversely with C. - **Parallel:** C_eq = C1 + C2; larger than the largest member; voltages equal, charges share in proportion to C. - **Battery disconnected versus connected:** isolated → Q constant, V → V0/K, U → U0/K; connected → V constant, Q → K Q0, U → K U0 — the single most exam-worthy contrast in the chapter. ## A slab slides in, twice Charge a 100 pF capacitor to 20 V and disconnect the battery: Q = 2 nC and U = (1/2) Q V = 2 × 10^-8 J, or 0.02 μJ. Now slide in a slab with K = 3, fully filling the gap. Capacitance triples to 300 pF; the charge has nowhere to go, so V = Q/C = 2 × 10^-9/3 × 10^-10 ≈ 6.7 V and U = Q^2/2C = (2 × 10^-9)^2/(6 × 10^-10) ≈ 6.7 × 10^-9 J — one-third of the original. The missing energy was spent pulling the slab in, because the slab is attracted into the gap. Repeat with the battery still connected: V stays pinned at 20 V, so Q rises to 6 nC and U to 0.06 μJ — tripled. The battery's ledger closes exactly: it pushes 4 nC of extra charge across 20 V, supplying 8 × 10^-8 J, of which 4 × 10^-8 J is newly stored energy and 4 × 10^-8 J is the work done drawing the slab in. The same slab therefore raises stored energy in a connected circuit and lowers it in an isolated one, and every sub-question about charge, field and voltage inherits its direction from that single distinction. ## The connected-or-not trap, and its cousins Nearly every slab question is decided before any arithmetic: identify whether Q or V is pinned — the wrong choice lands on decoy answers wrong by factors of K or K^2. The second trap is the energy-form choice: use (1/2) C V^2 when the battery fixes V, Q^2/2C when the capacitor is isolated — mixing them mid-solution is the most common carry-forward error in this chapter. Third, partial insertion: if a slab fills only half the gap, treat the arrangement as two capacitors in parallel (one air-filled, one dielectric-filled, same d) giving C = (ε0 A/2d)(1 + K) — the geometry, not the slab's presence alone, dictates the equivalent circuit. ## Frequently asked questions ### Why does inserting a dielectric increase capacitance? Polarisation charges oppose the applied field, lowering the internal field to E/K and the potential difference to V/K for the same charge, so C = Q/V rises by the factor K. ### What stays constant when a charged capacitor is disconnected from the battery? The charge Q — with no path to flow, it is pinned, and voltage, field and stored energy all fall as the dielectric is inserted. ### How do series and parallel capacitor combinations differ? Series adds reciprocals and forces equal charge with shared voltage; parallel adds directly, forcing equal voltage with shared charge. ### Where is the energy of a charged capacitor actually stored? In the electric field between the plates, with density u = (1/2) ε0 E^2 (multiplied by K for a dielectric filler). ### What happens to stored energy when a slab is inserted with the battery connected? It increases by the factor K along with the charge, with the battery supplying the additional energy — the opposite of the isolated case, where stored energy falls.