Dielectrics and Capacitance
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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.
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.