An air-filled parallel plate capacitor has a capacity $2 \mathrm{pF}$, The separation of the plates is…

An air-filled parallel plate capacitor has a capacity $2 \mathrm{pF}$, The separation of the plates is doubled and the interspace between the plates is filled with dielectric material, then the capacity is increased to $6 \mathrm{pF}$, The dielectric constant of the material is
  1. 3
  2. 6
  3. 2
  4. 4

Solution

$\begin{aligned} & \mathrm{C}=\frac{\mathrm{kA} \varepsilon_0}{\mathrm{~d}} \\ & \therefore \frac{\mathrm{C}_2}{\mathrm{C}_1}=\frac{\mathrm{k}_2}{\mathrm{k}_1} \cdot \frac{\mathrm{d}_1}{\mathrm{~d}_2} \\ & \therefore \frac{6}{2}=\frac{\mathrm{k}_2}{1} \cdot \frac{1}{2} \\ & \therefore \mathrm{k}_2=6 \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 1)

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