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Electrostatics accounts for approximately 6% of JEE Main marks. Success requires mastering both conceptual clarity and computational speed.

Strategy 1 — Symmetry Analysis: Before any calculation, examine the charge configuration for symmetry. Equal charges at polygon vertices, symmetric distributions, and infinite geometries all have predictable cancellation patterns. This can eliminate half the computation.

Strategy 2 — Method Selection: Point charges use Coulomb's law + superposition. Symmetric continuous distributions (sphere, cylinder, plane) use Gauss's law. Asymmetric distributions require direct integration. Mixing methods wastes time.

Strategy 3 — Dimensional Checking: Every intermediate and final answer should have correct dimensions. Force: [MLT^(-2)]. Field: [MLT^(-3)A^(-1)]. Flux: [$ML^{3T}$^(-3)A^(-1)]. A quick dimension check catches algebraic errors that waste time.

Strategy 4 — Limiting Cases: Verify that your answer reduces to known results. A sphere at large distance should give the point charge field. A disk of infinite radius should give the infinite plane result. A ring on its axis at x >> R should give kQ/$x^2$.

Strategy 5 — Distinguish E = 0 vs V = 0: These are NOT the same condition. E = 0 between two equal like charges, but V is not zero there. V = 0 at specific points between unequal opposite charges, but E is not zero there. JEE exploits this confusion.

Strategy 6 — Conductor vs Non-conductor: Inside a conductor, E = 0 always. Inside a non-conducting sphere with charge, E is proportional to r. This distinction is heavily tested.

Strategy 7 — Cavity Problems: For a non-conducting sphere with a cavity, use the superposition principle (full sphere minus removed sphere). The field inside the cavity is uniform: E = rho*$\frac{d}{3*epsilon_0}$, where d is the vector between centers. This elegant result appears every 2-3 years in JEE.