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Question

A parallel plate capacitor has plates of area \( 0.09 \, \text{m}^2 \) and separation 0.5 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

**Equipotential through midpoint** of +q and -q is perpendicular bisector, V=0 everywhere on it because contributions k q/r and k(-q)/r cancel. For two opposite charges, potential at midpoint zero, but field non-zero, pointing from positive to negative, illustrating vector vs scalar nature. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.09/0.5 × 10⁻³) = 1.593 × 10⁻⁹ F = 1593 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 1593 pF follows, reflecting potential-capacitance relations.

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