Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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
PhysicsCapacitanceCOMEDKCOMEDK 2017
Options:
  • A 2
  • B 3
  • C 4
  • D 6
Solution:
1022 Upvotes Verified Answer
The correct answer is: 6
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)]
$\Rightarrow \quad k=6$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.