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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance3 Marks
Question
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?
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Answer

Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
$\text{C}=\frac{\text{k}\in_0\text{A}}{\text{d}}$
$=\frac{\in_0\text{A}}{\text{d}} \dots\dots(1)$
Where,
A = Area of each plate
$∈_0=$ Permittivity of free space
If distance between the plates is reduced to half, then new distance, $\text{d}^{-\text{TM}}=\frac{\text{d}}{2}$ Dielectric constant of the substance filled in between the plates, k'= 6 Hence, capacitance of the capacitor becomes
$\text{C}'=\frac{\text{k}'\in_0\text{A}}{\text{d}}=\frac{6\in_0\text{A}}{\frac{\text{d}}{2}} \dots\dots(2)$
Taking ratios of equations (i) and (ii), we obtain
$C' = 2 × 6C$
$= 12\ C$
$= 12 × 8 = 96pF$
Therefore, the capacitance between the plates is 96 pF.

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