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#parallel plate capacitor

28 public questions tagged with this topic.

Displacement Current and Ampere-Maxwell Law

A parallel plate capacitor with plate area \( A = 0.01 \, \text{m}^2 \) and separation \( d = 2 \, \text{mm} \) is being

**Ampere-Maxwell law** ∮ B·dl = μ₀(I_c + ε₀ dΦ_E/dt) generalizes Ampere's law, displacement current arises from time-varying electric field, source of magnetic field like conduction current. For rate of change of flux 2×10¹¹ V·m/s, I_d = ε₀×2×10¹¹ =8.85×10⁻¹²×2×10¹¹=1.77 A. Displacement current i_d = ε₀ (d Φ_E/dt) . Since Φ_E = (Q/ε₀) , we have (d Φ_E/dt) = (1/ε₀) (dQ/dt) . But in a capacitor, i_d = (dQ/dt) . Thus, i_d = 0.5 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 0.5 A, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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Capacitor with Dielectric and Effect of Inserting Slab

A parallel plate capacitor with \( C = 90 \, \text{pF} \) in air has a dielectric (\( K = 9 \)) 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 = 9 × 90 = 810 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 810 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

When a dielectric slab fills only half the space between the plates of a parallel plate capacitor (connected to a batter

**Effect of dielectric** is to reduce effective field due to polarization, bound surface charges opposite to free charges, net field E = E₀ - E_p = E₀/K. Capacitance C = Q/V = K ε₀ A/d for full filling, partial filling C = ε₀ A/(d - t + t/K) for slab thickness t. With a constant voltage V across the plates, the electric field E varies between regions. In the air region, Eₐir = (V/d) , where d is the plate separation. In the dielectric region ( K > 1 ), polarization reduces the field: Ediₑlₑctric = (Eₐir/K) = (V/K d) . Since K > 1

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 dielectric slab is partially inserted between the plates of a parallel plate capacitor while maintaining a constant vo

**Effect of dielectric** is to reduce effective field due to polarization, bound surface charges opposite to free charges, net field E = E₀ - E_p = E₀/K. Capacitance C = Q/V = K ε₀ A/d for full filling, partial filling C = ε₀ A/(d - t + t/K) for slab thickness t. When a dielectric slab (with dielectric constant K > 1 ) is inserted between the plates of a capacitor with constant voltage V , the electric field E in the dielectric region decreases. The electric field in a dielectric is given by E = (E₀/K) , where E₀ = (V/d) is the field

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 capacitance \( 300 \, \text{pF} \) has a dielectric (\( K = 3 \), thickness \( d/8 \)) i

**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. Potential difference: V = E₀ ( (7d/8) ) + (E₀/K) ( (d/8) ) = E₀ d ( (7/8) + (1/8 × 3) ) . V = E₀ d ( (7/8) + (1/24) ) = E₀ d × (22/24) = E₀ d × (11/12) . C = (Q/V) = (Q/(11/12) V₀) = (12/11) × 300 ≈ 327.27 pF

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 \( C = 80 \, \text{pF} \) in air has a dielectric (\( K = 8 \)) 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 = 8 × 80 = 640 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 640 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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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

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Capacitor with Dielectric and Effect of Inserting Slab

A parallel plate capacitor with \( C = 20 \, \text{pF} \) in air has a dielectric (\( K = 3 \)) 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 = 3 × 20 = 60 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 60 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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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

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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

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Capacitor with Dielectric and Effect of Inserting Slab

A parallel plate capacitor with \( C = 40 \, \text{pF} \) in air has a dielectric (\( K = 5 \)) inserted fully between p

**Effect of dielectric** is to reduce effective field due to polarization, bound surface charges opposite to free charges, net field E = E₀ - E_p = E₀/K. Capacitance C = Q/V = K ε₀ A/d for full filling, partial filling C = ε₀ A/(d - t + t/K) for slab thickness t. C' = K C = 5 × 40 = 200 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 200 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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Capacitance of Parallel Plate and Spherical Capacitor

Why does the potential difference between the plates of a parallel plate capacitor remain constant when a dielectric sla

**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. When a capacitor is connected to a battery, the potential difference V across its plates is fixed by the battery. Inserting a dielectric slab (with K > 1 ) increases the capacitance ( C' = K C ), but the battery maintains V . To keep V constant ( Q = C V ), the charge Q on the plates increases ( Q' = C' V = K

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

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