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#electric field uniformity

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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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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

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