Q69 / 75 · VITEEE 2021 · Physics

A parallel plate capacitor with air between the plates has a capacitance of $8 \mathrm{pF}$. Calculate 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. (k =6)

  • A
    $72 \mathrm{pF}$
  • B
    $81 \mathrm{pF}$
  • C
    $84 \mathrm{pF}$
  • Correct answer
    D
    $96 \mathrm{pF}$

Solution

Capacity of parallel plate capacitor $$ \mathrm{C}=\frac{\mathrm{k} \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \quad\left(\text { For air } \mathrm{k}_{\mathrm{r}}=1\right) $$ So, $\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=8 \times 10^{-12}$ If $d \rightarrow \frac{d}{2}$ and $k_r \rightarrow 6$ then new capacitance $\mathrm{C}^{\prime}=6 \times \frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=12 \frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=12 \times 8 \mathrm{pF}=96 \mathrm{pF}$

Hence, the correct answer is option: 'D'.

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