Question

The capacitance of a parallel plate capacitor with air as medium is $3\mu F$. As a dielectric is introduced between the plates, the capacitance becomes$15\mu F$. The permittivity of the medium in \[{C^2}{N^{ - 1}}{m^{ - 2}}\;\] is:
A. $8.15 \times {10^{ - 11}}$
B. $0.44 \times {10^{ - 10}}$
C. $15.2 \times {10^{12}}$
D. $1.6 \times {10^{ - 14}}$

Answer 1

Hint To solve this question use the formula of capacitance between two parallel plates i.e. $c = {\dfrac{{A{ \in _0}}}{d}_{}}$ And when any dielectric is inserted between the capacitor plates, the capacitance increases by a factor of k i.e.
New capacitance ${c_d} = \dfrac{{kA{ \in _0}}}{d}$.Now, compare both the equations and use the formula of permittivity, that is $ \in = { \in _0}k$ and you will get the desired result

 Complete step-by-step solution:
Let us take the original capacitance be $'c'$ and the new capacitance, when dielectric is inserted between the plates, be $'{c_d}'$.
Using the formula of capacitance, $c = \dfrac{{A{ \in _0}}}{d} - - - (1)$
When dielectric is inserted between plates, the capacitance of capacitor increases by the factor of k (dielectric constant) i.e. the new formula is${c_d} = \dfrac{{kA{ \in _0}}}{d} - - - (2)$.
Where $A$area of capacitor is plates and $d$ is distance between the two parallel plates of the capacitor.
Now, as per the given question original capacitance $c = 3\mu F - - - (3)$
And, capacitance when dielectric is inserted is ${c_d} = 15\mu F - - - (4)$
Dividing equation $(1)$ and $(2)$, we get:
$\dfrac{c}{{{c_d}}} = \dfrac{{\dfrac{{A{ \in _0}}}{d}}}{{\dfrac{{kA{ \in _0}}}{d}}} - - - (5)$
$\dfrac{c}{{{c_d}}} = \dfrac{1}{k} - - - (6)$
Now, putting the values from equation $(3)$ and $(4)$in equation$(6)$:
We get,$\dfrac{{3\mu F}}{{15\mu F}} = \dfrac{1}{k}$
Solving this, we get $k = 5$
Now, to calculate the permittivity we will use the formula $ \in = { \in _0}k$
Where the value of ${ \in _0}$ is permittivity constant and it is \[{ \in _0} = 8.85 \times {10^{ - 12}}\]
When dielectric is inserted in between the capacitor plates, the permittivity constant also increased by a factor of $k$, so the new permittivity will be $ \in = { \in _0}k$
Putting the value of $k$ and ${ \in _0}$ , it becomes
\[ = 8.85 \times {10^{ - 12}} \times 5 = 0.44 \times {10^{ - 10}}\]


Therefore, the answer is option B.

Note: remember in such questions it is very obvious to make unit mistakes. So, always take care of units and ${ \in _0}$ is a constant, it’s value is fixed in every question and dielectric constant is different for every material, don’t get confused with the word constant.