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)

Capacitance in vacuum, $C_0=\frac{{\epsilon }_{0}A}{d}$
Capacitance in medium , $C_m={\epsilon }_0{\epsilon }_{\gamma }A/d$
1/8 = $V_m/V_0=(q_0/C)/(q/C_0)=C_0/C_m=1/{\epsilon }_r$
Hence ${\epsilon }_r=8$