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Question

A parallel plate capacitor has plates of area \( 0.06 \, \text{m}^2 \) and separation 0.3 mm in air.
What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1}
\text{m}^{-2} \)).

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Explanation

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.06/0.3 × 10⁻³) = 1.77 × 10⁻⁹ F = 1770 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 1770 pF follows, reflecting potential-capacitance relations.

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