Question

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 :
$$(\in_0= 8.85 \times 10^{-12} C^2N^{-1} m^{-2})$$

Answer 1

Correct option is B. $$0.44 \times 10^{-10} C^2 N^{-1} m^{-2}$$
Capacitance of air capacitor
$$C_0 = \dfrac{\varepsilon_0A}{d} = 6 \mu F$$ ......... (i)
When a dielectric of permittivity $$\varepsilon_r$$ and dielectric constant K is introduced between the plates, then
Capacitance, $$C = \dfrac{K \varepsilon_0 A}{d} = 30 \mu F$$ ..... (ii)
Dividing eq. (ii) by (i), we get
$$\dfrac{C}{C_0} = \dfrac{\dfrac{K\epsilon_o A}{ d}}{\dfrac{\varepsilon_0 A}{d}} = \dfrac{30}{6}$$
$$\Rightarrow K = 5$$
$$\therefore$$ permittivity of the medium
$$\varepsilon_0 = \varepsilon_0 K$$
$$= 8.85 \times 10^{-12} \times 5 = 0.44 \times 10^{-10}$$