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Question: The capacitance of a parallel plate capacitor with air as medium is \(3\mu F\). As a dielectric is i...

The capacitance of a parallel plate capacitor with air as medium is 3μF3\mu F. As a dielectric is introduced between the plates, the capacitance becomes 15μF15\mu F. The permittivity of the medium in C2N−1m−2  {C^2}{N^{ - 1}}{m^{ - 2}}\; is:
A. 8.15×10−118.15 \times {10^{ - 11}}
B. 0.44×10−100.44 \times {10^{ - 10}}
C. 15.2×101215.2 \times {10^{12}}
D. 1.6×10−141.6 \times {10^{ - 14}}

Explanation

Solution

Hint To solve this question use the formula of capacitance between two parallel plates i.e. c=A∈0dc = {\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 cd=kA∈0d{c_d} = \dfrac{{kA{ \in _0}}}{d}.Now, compare both the equations and use the formula of permittivity, that is ∈=∈0k \in = { \in _0}k and you will get the desired result

Complete step-by-step solution :
Let us take the original capacitance be ′c′'c' and the new capacitance, when dielectric is inserted between the plates, be ′cd′'{c_d}'.
Using the formula of capacitance, c=A∈0d−−−(1)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 cd=kA∈0d−−−(2){c_d} = \dfrac{{kA{ \in _0}}}{d} - - - (2).
Where AAarea of capacitor is plates and dd is distance between the two parallel plates of the capacitor.
Now, as per the given question original capacitance c=3μF−−−(3)c = 3\mu F - - - (3)
And, capacitance when dielectric is inserted is cd=15μF−−−(4){c_d} = 15\mu F - - - (4)
Dividing equation (1)(1) and (2)(2), we get:
ccd=A∈0dkA∈0d−−−(5)\dfrac{c}{{{c_d}}} = \dfrac{{\dfrac{{A{ \in _0}}}{d}}}{{\dfrac{{kA{ \in _0}}}{d}}} - - - (5)
ccd=1k−−−(6)\dfrac{c}{{{c_d}}} = \dfrac{1}{k} - - - (6)
Now, putting the values from equation (3)(3) and (4)(4) in equation(6)(6):
We get,3μF15μF=1k\dfrac{{3\mu F}}{{15\mu F}} = \dfrac{1}{k}
Solving this, we get k=5k = 5
Now, to calculate the permittivity we will use the formula ∈=∈0k \in = { \in _0}k
Where the value of ∈0{ \in _0} is permittivity constant and it is ∈0=8.85×10−12{ \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 kk, so the new permittivity will be ∈=∈0k \in = { \in _0}k
Putting the value of kk and ∈0{ \in _0} , it becomes
=8.85×10−12×5=0.44×10−10= 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 ∈0{ \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.