Practice question
Question
In a charged parallel plate capacitor, why does the electric field remain uniform between the plates
even when a dielectric slab is partially inserted?
Explanation
**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
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