Question

The capacitance of a parallel plate capacitor with air as medium is $6\mu \text{F}$. With the introduction of a dielectric medium, the capacitance becomes $30\mu \text{F}$. The permittivity of the medium is: ${\epsilon }_0=8.85\times {10}^{-12}{\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$

• A. $0.44\times {10}^{-10}{\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$
• B. $5.00{\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$
• C. $0.44\times {10}^{-13}{\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$
• D. $1.77\times {10}^{-12}{\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$

Answer 1

The correct option is D $1.77\times {10}^{-12}{\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$

Given,
$C_{air}=6\mu \text{F}$, $C_{medium}=30\mu \text{F}$
Let dielectric constant of medium be $k$.
We know, $C_{medium}=k{C}_{air}$
$\Rightarrow 30=k\times 6$
$\therefore k=5$
Now, permittivty of the medium, $\mu ={\mu }_{0}k$
$\Rightarrow \mu =8.85\times {10}^{-12}\times 5$
$\therefore \mu =0.44\times {10}^{-10}{\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$