#plate area

Conductors, Electrostatic Shielding and Dielectrics

A parallel plate capacitor has plates of area \( 0.07 \, \text{m}^2 \) and separation 0.35 mm in air. What is its capaci

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.07/0.35 × 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.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

View Details

Conductors, Electrostatic Shielding and Dielectrics

A parallel plate capacitor has plates of area \( 0.05 \, \text{m}^2 \) and separation 0.25 mm in air. What is its capaci

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.05/0.25 × 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.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

View Details

Equipotential Surfaces and Relation Between Field and Potential

A parallel plate capacitor has plates of area \( 0.06 \, \text{m}^2 \) and separation 0.3 mm in air. What is its capacit

**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.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Equipotential Surfaces and Relation Between Field and Potential

View Details

Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor has plates of area \( 0.05 \, \text{m}^2 \) separated by 1 mm in air. What is its capacitance

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.05/10⁻³) = 4.425 × 10⁻¹⁰ F = 442.5 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 442.5 pF follows, reflecting potential-capacitance relations.