Search any question & find its solution
Question:
Answered & Verified by Expert
If a slab of insulating material (conceptual). $4 \times 10^{-3} \mathrm{~m}$ thick is introduced between the plates of a parallel plate capacitor, the separation between the plates has to be increased by $3.5 \times 10^{-3} \mathrm{~m}$ to restore the capacity to original value. The dielectric constant of the material will be
Options:
Solution:
2446 Upvotes
Verified Answer
The correct answer is:
8
Let $t$ be the thickness of the dielectric slab and $K$ is the dielectric constant.
So, the increase in the distance of separation between the plates due to dielectric is given as
$\begin{aligned}
x &=t-\frac{t}{K} \\
&=t\left(1-\frac{1}{k}\right)
\end{aligned}$
Given, $x=3.5 \times 10^{-3} \mathrm{~m}, t=4 \times 10^{-3} \mathrm{~m}$
Substituting the given values in the above equation, we get
$1-\frac{1}{k}=\frac{x}{t}=\frac{3.5 \times 10^{-3}}{4 \times 10^{-3}}=\frac{3.5}{4}$
or
$\frac{1}{k}=1-\frac{3.5}{4}=\frac{0.5}{4}$
or
$k=8$
So, the increase in the distance of separation between the plates due to dielectric is given as
$\begin{aligned}
x &=t-\frac{t}{K} \\
&=t\left(1-\frac{1}{k}\right)
\end{aligned}$
Given, $x=3.5 \times 10^{-3} \mathrm{~m}, t=4 \times 10^{-3} \mathrm{~m}$
Substituting the given values in the above equation, we get
$1-\frac{1}{k}=\frac{x}{t}=\frac{3.5 \times 10^{-3}}{4 \times 10^{-3}}=\frac{3.5}{4}$
or
$\frac{1}{k}=1-\frac{3.5}{4}=\frac{0.5}{4}$
or
$k=8$
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.