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#battery connection

2 public questions tagged with this topic.

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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Capacitor with Dielectric and Effect of Inserting Slab

Why does the capacitance of a parallel plate capacitor increase when a dielectric slab is inserted between the plates wh

**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 while the capacitor remains connected to a battery (constant voltage V ), the electric field decreases due to polarization ( E = E₀/K ), and the capacitance increases ( C' = K C ). This happens because the dielectric reduces the field,

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

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