#dielectric slab

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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Capacitance of Parallel Plate and Spherical Capacitor

In a charged parallel plate capacitor, why does the electric field remain uniform between the plates even when a dielect

**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. The electric field between the plates of a parallel plate capacitor is ideally uniform ( E = (sigma/ε₀) ) in air. When a dielectric slab is partially inserted, the field in the air region remains E₀ = (sigma/ε₀) , and in the dielectric region, it reduces to E = (E₀/K) . However, within each region (air or dielectric), the field remains uniform because the plates are large and

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

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Capacitor with Dielectric and Effect of Inserting Slab

In a parallel plate capacitor with a partially inserted dielectric slab while maintaining constant charge, why does the

**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. Capacitance C = (Q/V) . With constant charge Q , inserting a dielectric ( K > 1 ) increases C by reducing V ( C = (K ε₀ A/d) ). As the slab is withdrawn, the effective dielectric constant decreases (less area has K ), lowering C . Since Q is constant, V

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 inserted into a parallel plate capacitor while maintaining a constant charge. Why does the potentia

**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. With a constant charge Q , capacitance C = (ε₀ A/d) without dielectric, and C' = (K ε₀ A/d) with dielectric ( K > 1 ). When the dielectric is fully inserted, C increases, reducing V = (Q/C) . If the slab is partially removed, the effective capacitance decreases (as less dielectric area contributes K ),

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

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