A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
The new capacitance is 96 pF.
Explanation
This question tests the understanding of how capacitance depends on geometry and dielectric material. From the context, the capacitance of a parallel plate capacitor is C₀ = ε₀A/d (Eq. 2.43). When a dielectric is introduced, the capacitance becomes C = KC₀, where K is the dielectric constant. The question combines two effects: reducing the distance by half (which doubles capacitance since C ∝ 1/d) and introducing a dielectric (which multiplies capacitance by K). The combined effect gives C_new = 2K × C₀ = × 8 pF = 96 pF.
Solution Steps
Step 1: Write the formula for capacitance of parallel plate capacitor with air: C₀ = ε₀A/d = 8 pF
Step 2: When distance is reduced by half (d' = d/2), capacitance becomes: C' = ε₀A/(d/2) = 2ε₀A/d = 2C₀ = = 16 pF
Step 3: When dielectric of constant K = 6 is filled, capacitance becomes: C = K × C' = = 96 pF
Alternative method: C_new = 2K × C₀ = × 8 = 96 pF