A parallel plate capacitor with air between the plates has a capacitance of 4 pF. If the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 6 then the value of capacitance will be .

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(4) 48 pF. Given: C₀ = (ε₀A)/d = 4 pF; new gap d/2, filled with a dielectric of constant K = 6. Idea: C = (Kε₀A)/(d') — a dielectric multiplies C by K, and C is inversely proportional to the gap. C = (6 ε₀A)/(d/2) = 12 (ε₀A)/d = 12C₀. C = 12×4 = 48 pF. So the capacitance becomes 48 pF.

A parallel plate capacitor with air between the plates has a capacitance of 4 pF. If the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 6 then the value of capacitance will be ________.

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