#parallel plate

Production of EM Waves and Hertz Experiment

A parallel plate capacitor with plate area \( A = 0.02 \, \text{m}^2 \) and separation \( d = 5 \, \text{mm} \) is conne

**Production of EM waves** requires accelerated charge, oscillating LC circuit produces changing E and B, antenna radiates when charge accelerates, frequency determined by L and C, f=1/(2π√(LC)). Hertz used spark gap with inductor and capacitor, produced ~10⁸ Hz radio waves, detected with loop, confirmed transverse nature, reflection, refraction, polarization, speed c. Displacement current i_d = ε₀ (d Φ_E/dt) . For a capacitor, i_d = i . Given i = 2 A , we have (d Φ_E/dt) = (i/ε₀) = (2/8.85 × 10⁻¹²) ≈ 2.26 × 10¹¹ Vm/s . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f,

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

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Capacitor with Dielectric and Effect of Inserting Slab

A parallel plate capacitor with \( C = 70 \, \text{pF} \) in air has a dielectric (\( K = 5 \)) inserted fully between p

**Potential difference increase** statement with constant charge is incorrect; V decreases when K>1 inserted. If slab inserted while maintaining constant charge, E reduces to E₀/K, V = E d reduces. If battery connected maintaining constant V, E stays V/d, D = ε E increases, Q increases K times. C' = K C = 5 × 70 = 350 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 350 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitor with Dielectric and Effect of Inserting Slab

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Capacitor with Dielectric and Effect of Inserting Slab

A parallel plate capacitor with \( C = 30 \, \text{pF} \) in air has a dielectric (\( K = 6 \)) inserted fully between p

**Dielectric slab inserted** into capacitor with constant charge Q increases capacitance C' = K C₀, K dielectric constant, so potential difference V' = Q/C' = V₀/K decreases K times. With constant voltage V maintained by battery, capacitance increase causes charge Q' = K Q₀ to increase, extra charge supplied by battery, energy changes. C' = K C = 6 × 30 = 180 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 180 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitor with Dielectric and Effect of Inserting Slab

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Equipotential Surfaces and Relation Between Field and Potential

A parallel plate capacitor has plates of area \( 0.09 \, \text{m}^2 \) and separation 0.5 mm in air. What is its capacit

**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.

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

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 50 \, \text{pF} \) has a dielectric (\( K = 2 \), thickness \( d/3 \)) in

**Parallel plate capacitor** capacitance C = ε₀ A/d, ε₀=8.85×10⁻¹² F/m, A plate area (m²), d separation (m), for air, with dielectric C = K ε₀ A/d. For A=0.08 m², d=0.4 mm=4×10⁻⁴ m, C=8.85×10⁻¹²×0.08/4×10⁻⁴=1.77×10⁻⁹ F=1.77 nF, illustrating small capacitance for cm separation. Potential difference: V = E₀ ( (2d/3) ) + (E₀/K) ( (d/3) ) = E₀ d ( (2/3) + (1/3 × 2) ) = E₀ d ( (2/3) + (1/6) ) = E₀ d × (5/6) . C = (Q/V) = (Q/(5/6) V₀) = (6/5) × 50 = 60 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =

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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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor has plates of area \( 0.03 \, \text{m}^2 \) and separation 0.5 mm in air. What is its capacit

**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 × 0.03/0.5 × 10⁻³) = 5.31 × 10⁻¹⁰ F = 531 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 531 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 has plates of area \( 0.04 \, \text{m}^2 \) and separation 0.2 mm in air. What is its capacit

**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 = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.04/0.2 × 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 > Capacitance of Parallel Plate and Spherical Capacitor

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