Q.Define capacitance of a capacitor and its SI unit.

Capacitance measures how much charge a conductor arrangement stores for every volt you push across it: C = Q/V. The subtle point JEE Main keeps testing is that C is a geometric property — it depends only on the shape, size, and spacing of the conductors and the material between them, not on Q or V. Put more charge on and V rises in exact proportion, leaving the ratio fixed. So C is a container's "capacity," decided the moment the container is built.

Why parallel plates give C = ε₀A/d. For two plates of area A a distance d apart, the field between them is uniform, E = σ/ε₀ = Q/(ε₀A), and the voltage is V = Ed = Qd/(ε₀A). Dividing, C = ε₀A/d. Bigger plates hold more charge at the same voltage (C ∝ A), while pulling the plates closer raises C (C ∝ 1/d) because a smaller gap means less voltage is needed to hold the same charge.

Series vs parallel — reason it out, don't memorise. In series the same charge Q sits on every capacitor (the isolated middle plates must stay neutral), so the voltages add: V = Q/C₁ + Q/C₂ + …, giving 1/C_eq = Σ 1/Cᵢ. Series capacitance is smaller than the smallest member. In parallel every capacitor shares the same voltage V, so the charges add: Q = C₁V + C₂V + …, giving C_eq = Σ Cᵢ. Effectively you have enlarged the plate area.

Dielectrics. Slip an insulator of dielectric constant K between the plates and C rises to KC₀. The molecules polarise, setting up an internal field that opposes the applied one, so the net field — and hence the voltage per unit charge — drops by a factor K. More charge now fits at the same voltage.

The decisive question: is the battery still connected? This is where marks are won or lost when a dielectric is inserted. …