#dielectric constant

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

View Details

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

View Details

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

View Details

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

View Details

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

View Details

Conductors, Electrostatic Shielding and Dielectrics

Why does the electric field inside a dielectric material decrease when placed in an external field, compared to the fiel

**Conductor in electrostatic equilibrium** has E=0 inside, charges reside on surface, potential constant throughout conductor. Hollow shell with no internal charge has zero field inside cavity, even if external field present, charges on outer surface screen interior, principle used in Faraday cage. When a dielectric is placed in an external field E₀ , it polarizes, creating bound charges that produce an internal field Eiₙducₑd opposing E₀ . The net field inside the dielectric is E = (E₀/K) , where K > 1 is the dielectric constant. For linear dielectrics, K > 1 , so E < E₀ , as the polarization reduces the effective field

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

View Details

Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with \( C = 50 \, \text{pF} \) in air has a dielectric (\( K = 4 \)) inserted fully between p

**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. C' = K C = 4 × 50 = 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 > Capacitance of Parallel Plate and Spherical Capacitor

View Details

Combination of Capacitors - Series and Parallel

In a series combination of capacitors with different dielectric materials between their plates, why do capacitors with h

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. In series, the charge Q on each capacitor is the same. Capacitance is C = (K ε₀ A/d) , where K is the dielectric constant. A higher K increases C . Since V = (Q/C) , a larger C (due to higher K ) results in a smaller V . Thus, capacitors with higher dielectric constants

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

View Details

Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 120 \, \text{pF} \) has a dielectric (\( K = 4 \), thickness \( d/4 \)) 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 ( (3d/4) ) + (E₀/K) ( (d/4) ) = E₀ d ( (3/4) + (1/4 × 4) ) . V = E₀ d ( (3/4) + (1/16) ) = E₀ d × (13/16) . C = (Q/V) = (Q/(13/16) V₀) = (16/13) × 120 ≈ 147.69 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

View Details

Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with \( C = 100 \, \text{pF} \) in air has a dielectric (\( K = 10 \)) inserted fully between

**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 = 10 × 100 = 1000 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 1000 pF follows, reflecting potential-capacitance relations.

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

View Details

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

View Details

Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 250 \, \text{pF} \) has a dielectric (\( K = 4 \), thickness \( d/7 \)) i

**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. Potential difference: V = E₀ ( (6d/7) ) + (E₀/K) ( (d/7) ) = E₀ d ( (6/7) + (1/7 × 4) ) . V = E₀ d ( (6/7) + (1/28) ) = E₀ d × (25/28) . C = (Q/V) = (Q/(25/28) V₀) = (28/25) × 250 = 280 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

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

View Details