Electrostatic Potential And Capacitance Ques 17
17. 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 : $(\in_0=8.85 \times 10^{-12}$ $ C^{2} N^{-1} m^{-2})$
[2020]
(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} $ $C^{2} N^{-1} m^{-2}$
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Answer:
Correct Answer: 17.(b)
Solution:
- (b) Capacitance of a parallel plate capacitor with air is
$C=\frac{\varepsilon_0 A}{d} $ $\quad$ …….(i)
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
$C^{\prime}=\frac{K \varepsilon_0 A}{d} $ $\quad$ …….(ii)
Dividing (ii) by (i)
$ \Rightarrow \frac{C^{\prime}}{C}=K \Rightarrow \frac{30}{6}=K \Rightarrow K=5 $
$\Rightarrow K=\frac{\varepsilon}{\varepsilon_0}$
$\Rightarrow \varepsilon=K \varepsilon_0=5 \times 8.85 \times 10^{-12}$
= $0.44 \times 10^{-10} $ $C^{2} N^{-1} m^{-2}$
BETA
