Electrostatic Potential and Capacitance - Result Question 22
24. The capacitance of a parallel plate capacitor with air as medium is $6 \mu F$. With the introduction of a dielectric medium, the capacitance becomes $30 \mu F$. The permittivity of the medium is :
[2020]
$ (\in_0=8.85 \times 10^{-12} C^{2} N^{-1} m^{-2}) $
(a) $1.77 \times 10^{-12} C^{2} N^{-1} m^{-2}$
(b) $0.44 \times 10^{-10} C^{2} N^{-1} m^{-2}$
(c) $5.00 C^{2} N^{-1} m^{-2}$
(d) $0.44 \times 10^{-13} C^{2} N^{-1} m^{-2}$
Show Answer
Answer:
Correct Answer: 24. (b)
Solution:
- (b) Capacitance of a parallel plate capacitor with air is
$ \begin{equation*} C=\frac{\varepsilon_0 A}{d} \tag{i} \end{equation*} $
Here, $A=$ area of plates of capacitor,
$d=$ distance between the plates
Capacitance of a same parallel plate capacitor with introduction of dielectric medium of dielectric constant $K$ is
$ \begin{equation*} C^{\prime}=\frac{K \varepsilon_0 A}{d} \tag{ii} \end{equation*} $
Dividing (ii) by (i)
$ \Rightarrow \frac{C^{\prime}}{C}=K \Rightarrow \frac{30}{6}=K \Rightarrow K=5 $
$\Rightarrow K=\frac{\varepsilon}{\varepsilon_0}$
$\Rightarrow \varepsilon=K \varepsilon_0=5 \times 8.85 \times 10^{-12}$
