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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 and constants : C = A / . For parallel with areas : C = sum of */d.
Conducting slab insertion: Equivalent to reducing the gap by the slab thickness. C = *. If the slab touches one plate, it effectively moves that plate.
Force on dielectric (battery on): F = (K-1)*wV^, constant, directed inward. This can be derived from F = (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 = ). 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.