#parallel plates

Capacitor with Dielectric and Effect of Inserting Slab

When a dielectric slab is inserted between the plates of a charged parallel plate capacitor (disconnected from the batte

**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 K > 1 ) is inserted into a disconnected capacitor, the charge Q remains constant. The capacitance increases ( C' = K C ), and the potential difference decreases ( V' = V/K ). The energy stored is given by U = (Q²/2C) , so with increased C

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

View Details

Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 150 \, \text{pF} \) has a dielectric (\( K = 6 \), thickness \( d/5 \)) 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 ( (4d/5) ) + (E₀/K) ( (d/5) ) = E₀ d ( (4/5) + (1/5 × 6) ) . V = E₀ d ( (4/5) + (1/30) ) = E₀ d × (25/30) = E₀ d × (5/6) . C = (Q/V) = (Q/(5/6) V₀) = (6/5) × 150 = 180 pF . Using V = kQ/r, U = k q₁q₂/r, E =

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

View Details

Capacitance of Parallel Plate and Spherical Capacitor

Why does the electric field between two large parallel plates with opposite charges remain uniform even if one plate has

**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. For large parallel plates with charges +sigma and -sigma , the field between them is E = (sigma/ε₀) , uniform due to symmetry. A dielectric coating on one plate polarizes, inducing bound charges, but does not alter the free charge on the plates. The net field between the plates remains determined by the free charge density sigma , and the unif

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 \( 60 \, \text{pF} \) has a dielectric (\( K = 4 \), thickness \( d/4 \)) 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 ( (3d/4) ) + (E₀/K) ( (d/4) ) = E₀ d ( (3/4) + (1/4 × 4) ) = E₀ d ( (3/4) + (1/16) ) = E₀ d × (13/16) . C = (Q/V) = (Q/(13/16) V₀) = (16/13) × 60 ≈ 73.85 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