Parallel-Plate CapacitorJEE Main

Concept

A capacitor stores charge — and energy — in the electric field between two plates. Its capacitance depends only on geometry and the dielectric filling the gap, not on the voltage applied.

Key formula

Derivation

A uniform field $E = V/d$ between plates of area $A$ holds charge $Q = \varepsilon_0\varepsilon_r A E$. Dividing by $V$ gives $C = \varepsilon_0\varepsilon_r A/d$.

Bringing the plates closer (smaller $d$), enlarging them, or inserting a dielectric ($\varepsilon_r > 1$) all increase $C$. The stored energy $U = \tfrac12 CV^2 = \tfrac12 QV$ lives in the field itself.

Scenarios to explore

• Parallel-Plate Capacitor — Capacitance, charge, energy & dielectrics.

Real-world applications

• Energy storage (camera flashes, power smoothing).
• Tuning circuits and touchscreens.
• Memory cells in DRAM.

JEE exam tips

• Series capacitors add reciprocals; parallel ones add directly (opposite to resistors).
• Inserting a dielectric at constant charge lowers $V$ and the stored energy; at constant voltage it raises both $Q$ and $U$.

Common mistakes

• Thinking $C$ depends on $Q$ or $V$ — it depends only on geometry and dielectric.
• Forgetting unit conversions (cm²→m², mm→m).
• Confusing energy $\tfrac12 CV^2$ with charge $CV$.

Exam traps to avoid

• Halving the plate gap doubles the capacitance and the stored energy (at fixed V).
• The energy density in the field is $u = \tfrac12\varepsilon_0\varepsilon_r E^2$.