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#capacitance change

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with \( C = 60 \, \text{pF} \) in air has a dielectric (\( K = 7 \)) inserted fully between p

**Capacitance depends on geometry** not charge, C = Q/V constant for given arrangement. Parallel plate C ∝ A/d, so larger area and smaller separation increase capacitance, principle used to increase storage by using large area foils with thin dielectric. C' = K C = 7 × 60 = 420 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 420 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 200 \, \text{pF} \) has a dielectric (\( K = 5 \), thickness \( d/6 \)) i

**Capacitance depends on geometry** not charge, C = Q/V constant for given arrangement. Parallel plate C ∝ A/d, so larger area and smaller separation increase capacitance, principle used to increase storage by using large area foils with thin dielectric. Potential difference: V = E₀ ( (5d/6) ) + (E₀/K) ( (d/6) ) = E₀ d ( (5/6) + (1/6 × 5) ) . V = E₀ d ( (5/6) + (1/30) ) = E₀ d × (26/30) = E₀ d × (13/15) . C = (Q/V) = (Q/(13/15) V₀) = (15/13) × 200 ≈ 230.77 pF . Using V = kQ/r, U = k q₁q₂/r,

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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