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#constant voltage

2 public questions tagged with this topic.

A dielectric slab is partially inserted between the plates of a parallel plate capacitor while maintaining a constant vo

**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 dielectric constant K > 1 ) is inserted between the plates of a capacitor with constant voltage V , the electric field E in the dielectric region decreases. The electric field in a dielectric is given by E = (E₀/K) , where E₀ = (V/d) is the field

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

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