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Question: The capacitance of a parallel plate capacitor with air as the medium is \( 6\,\mu F \) .With the int...

The capacitance of a parallel plate capacitor with air as the medium is 6 μF6\,\mu F .With the introduction of a dielectric medium, the capacitance becomes 30 μF30\,\mu F . The permittivity of the medium is ( ε0=8.85×10−12 C2 N−1 m−2 {\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}\,{N^{ - 1}}\,{m^{ - 2}}\, )
a) 1.77×10−12 C2 N−1 m−2 1.77 \times {10^{ - 12}}\,{C^2}\,{N^{ - 1}}\,{m^{ - 2}}\,
b) 0.44×10−10 C2 N−1 m−2 0.44 \times {10^{ - 10}}\,{C^2}\,{N^{ - 1}}\,{m^{ - 2}}\,
c) 5.00 C2 N−1 m−2 5.00\,{C^2}\,{N^{ - 1}}\,{m^{ - 2}}\,
d) 0.44×10−13 C2 N−1 m−2 0.44 \times {10^{ - 13}}\,{C^2}\,{N^{ - 1}}\,{m^{ - 2}}\,

Explanation

Solution

Hint : Capacitance is described as the property to hold charge. Capacitance is dependent on the geometry of the capacitor.
The dielectric constant is the ratio of permittivity of a material to the permittivity of free space. We will first use the formulas for the capacitance in the absence and presence of a dielectric medium and take their ratio to obtain a simpler expression in terms of the capacitances and the dielectric constant. Then we will use our fundamental knowledge that the dielectric constant is the ratio of permittivity of a material to the permittivity of free space and solve further to get the answer.
The mathematical expression for a parallel plate capacitor is C0=Aε0d{C_0} = \dfrac{{A{\varepsilon _0}}}{d} where C is the capacitance, A is the area of cross section of the capacitor, d is the separation between the plates and ε0{\varepsilon _0} is the permittivity of free space. This is the expression for the capacitor with no dielectric in between.
When a dielectric of dielectric constant K is present between the plates, the capacitance is given by Ck=KAε0d{C_k} = \dfrac{{KA{\varepsilon _0}}}{d} .

Complete Step By Step Answer:
We know that the dielectric constant is the ratio of permeability of a material to the permeability of free space. The expression is given as K=εrε0K = \dfrac{{{\varepsilon _r}}}{{{\varepsilon _0}}} .
Let the capacitance in a dielectric medium be Cr{C_r} . Its value is given by Cr=Aεrd{C_r} = \dfrac{{A{\varepsilon _r}}}{d} which simplifies to Cr=KAε0d{C_r} = \dfrac{{KA{\varepsilon _0}}}{d} .
But we know that the value of capacitance with the same geometry but in free space will be C0=Aε0d{C_0} = \dfrac{{A{\varepsilon _0}}}{d} .
Taking the ratio of Cr{C_r} and C0{C_0} ,
CrC0=KAε0dAε0d\dfrac{{{C_r}}}{{{C_0}}} = \dfrac{{\dfrac{{KA{\varepsilon _0}}}{d}}}{{\dfrac{{A{\varepsilon _0}}}{d}}}
Further solving this equation, we get
CrC0=K\dfrac{{{C_r}}}{{{C_0}}} = K where K is the dielectric constant of the medium.
Now we know that the dielectric constant is the ratio of permittivity of a material to the permittivity of free space.
K=εkε0K = \dfrac{{{\varepsilon _k}}}{{{\varepsilon _0}}}
Hence, we can say that CrC0=K=εkε0\dfrac{{{C_r}}}{{{C_0}}} = K = \dfrac{{{\varepsilon _k}}}{{{\varepsilon _0}}}
We know that C0=6 μF{C_0} = 6\,\mu F , Ck=30 μF{C_k} = 30\,\mu F , ε0=8.85×10−12 C2 N−1 m−2 {\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}\,{N^{ - 1}}\,{m^{ - 2}}\,
Putting the known values,
306=εk8.85×10−12 \dfrac{{30}}{6} = \dfrac{{{\varepsilon _k}}}{{8.85 \times {{10}^{ - 12}}\,}}
Further solving,
⇒5=εk8.85×10−12 \Rightarrow 5 = \dfrac{{{\varepsilon _k}}}{{8.85 \times {{10}^{ - 12}}\,}}
⇒εk=0.44×10−10 C2 N−1 m−2 \Rightarrow {\varepsilon _k} = 0.44 \times {10^{ - 10}}\,{C^2}\,{N^{ - 1}}\,{m^{ - 2}}\,
Hence, option B is the correct answer.

Note :
Capacitance is dependent on its geometry and this formula is applicable only for parallel plate capacitors. The expression is different for spherical capacitors which is dependent on the radius of the concentric. If nothing is mentioned, we take the capacitor to be a parallel plate capacitor and accordingly apply the formulas.