Part of JES-02 — Electrostatic Potential, Capacitance & Energy

Special Capacitor Problems

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Several capacitor problem types recur in JEE with specific solution strategies.

Partially filled dielectrics: Model as series (layers perpendicular to E) or parallel (sections dividing plate area). For series with thicknesses tit_i and constants KiK_i: C = epsilon0epsilon_0A / sumoftiKi\frac{sum of t_i}{K_i}. For parallel with areas AiA_i: C = sum of KiK_iepsilon0epsilon_0*AiA_i/d.

Conducting slab insertion: Equivalent to reducing the gap by the slab thickness. C = epsilon0epsilon_0*Adt\frac{A}{d-t}. If the slab touches one plate, it effectively moves that plate.

Force on dielectric (battery on): F = (K-1)*epsilon0epsilon_0wV^22d\frac{2}{2d}, constant, directed inward. This can be derived from F = dUdx\frac{dU}{dx} (keeping V constant and accounting for battery work).

Capacitor with varying separation: If plates are not parallel (wedge-shaped gap), divide into infinitesimal parallel strips and integrate. This creates a non-uniform capacitor.

Spherical shell problems: For concentric shells with charges, find the potential at each shell by summing contributions from all shells. Grounding a shell sets its potential to zero, which determines the charge redistribution.

Connected spheres: When two conducting spheres are connected by a wire, charge flows until both reach equal potential (V = kQR\frac{kQ}{R}). Charge distributes in proportion to radii: Q proportional to R. Surface charge density sigma is inversely proportional to R — smaller spheres have higher sigma and higher E at the surface.

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