Question

The capacitance of a parallel plate capacitor is $5\,\mu F$. When a glass slab is placed between the plates of the capacitor, its potential difference reduces to $1/8$ of the original value. The thickness of the slab is equal to the separation between the plates. The value of the dielectric constant of glass is

• A. 1.6
• B. 5
• C. 8
• D. 40

Answer 1

Answer: (C)

8

Answer 2

Capacitance in vacuum, $C_0 = \frac{\varepsilon_0 \, A}{d}$ Capacitance in medium , $C_m = \varepsilon_0 \, \varepsilon_{\gamma} A/d$ 1/8 = $V_m / V_0 = (q_0 / C) / ( q/C_0) = C_0 / C_m = 1/ \varepsilon_r$ Hence $\varepsilon_r = 8$