1. Tardigrade
  2. Question
  3. Physics
  4. The capacitance of a parallel plate capacitor is 5 μ 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

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Q. 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

Electrostatic Potential and Capacitance

Solution:

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 $\varepsilon_r = 8$