Question
A parallel plate capacitor with air between its plates has a capacitance of 8 pF(1 pF = 10-12F) . 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?

Answer

(i) Here is a parallel plate capacitor, which has air between its plates.
$C_1=\frac{\in_0 A}{d}=8 pF =8 \times 10^{-12} F$
(ii) When a substance of dielectric constant 6 is filled in the space between the plates
$C _2= K \frac{\in_0 A}{d^{\prime}}$
Here it is given K = 6 d' = d / 2
$C _2=\frac{6\left(\in_0 A\right)}{d / 2}=2 \times 6 \times\left(\frac{\in_0 A}{d}\right)$
$C_2=12 C_1$
but $C_1=8 \times 10^{-12} F$
Hence on putting values
$C _2=12 \times 8 \times 10^{-12} F$
$=96 \times 10^{-12} F=96 pF$

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