Q165 / 180 · COMEDK 2017 · Physics
An air filled parallel plate condenser has a capacity of $2 \mathrm{pF}$. The separation of the plates is doubled and the interspace between the plates is filled with wax. If the capacity is increased to $6 \mathrm{pF}$, then the dielectric constant of wax is
- A2
- B3
- C4
- Correct answerD6
Solution
Given, capacitance of air filled parallel plate capacitor,
$C_{0}=2 \mathrm{pF}$
$\Rightarrow \quad=2 \times 10^{-12} \mathrm{~F}$
$\Rightarrow \quad \frac{\varepsilon_{0} A}{d}=2 \times 10^{-12} \mathrm{~F}...(i)$
When the separation between the plates is doubled, i.e., $\quad d^{\prime}=2 d$
And wax of dielectric constant $k$ is filled between the plates of capacitor, then,
$\begin{aligned}
C^{\prime} &=6 \mathrm{pF} \\
\frac{\varepsilon_{0} k A}{d^{\prime}} &=6 \times 10^{-12} \mathrm{~F}
\end{aligned}$
$\Rightarrow \quad \frac{\varepsilon_{0} k A}{2 d}=6 \times 10^{-12} \quad\left[\because d^{\prime}=2 d\right]$
$\Rightarrow \quad \frac{\varepsilon_{0} A}{d} \times \frac{k}{2}=6 \times 10^{-12}$
$\Rightarrow 2 \times 10^{-12} \times \frac{k}{2}=6 \times 10^{-12}$
[from $\mathrm{Eq} .$ (i)]
$\Rightarrow \quad k=6$
Hence, the correct answer is option: 'D'.