The capacitance of a parallel plate capacitor with air as medium is 6μF. With the introduction of a dielectric medium, the capacitance becomes 30μF. The permittivity ?

The capacitance of a parallel plate capacitor with air as medium is 6μF. With the introduction of a dielectric medium, the capacitance becomes 30μF. The permittivity ?

\( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \, \text{N}^{-1} \, \text{m}^{-2} \)

(a) \( 1.77 \times 10^{-12} \, \text{C}^2 \, \text{N}^{-1} \, \text{m}^{-2} \)

(b) \( 0.44 \times 10^{-10} \, \text{C}^2 \, \text{N}^{-1} \, \text{m}^{-2} \)

(c) \( 5.00 \, \text{C}^2 \, \text{N}^{-1} \, \text{m}^{-2} \)

(d) \( 0.44 \times 10^{-13} \, \text{C}^2 \, \text{N}^{-1} \, \text{m}^{-2} \)

Answer :  (b) \( 0.44 \times 10^{-10} \, \text{C}^2 \, \text{N}^{-1} \, \text{m}^{-2} \)

\( C_0 = 6\, \mu F \)

\( C = K\, C_0 \)

\( 30\, \mu = K \cdot 6\, \mu \)

\( K = \varepsilon_r = \frac{\varepsilon_{\text{medium}}}{\varepsilon_0} \)

\( \varepsilon_{\text{medium}} = \varepsilon_r \cdot \varepsilon_0 \)

\( = 5 \times \left( 8.85 \times 10^{-12} \right) \)

\( = 44 \times 10^{-12} \)

\( = 0.44 \times 10^{-10} \)